Fangwen Shu
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.DS_Store
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.DS_Store
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*.txt
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!tencent13147342467085209222.txt
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desktop.ini
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README.md
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<div align="center">
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<img src="other/LogoMakr_0zpEzN.png" width="200px">
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<br>
|
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<a href="https://cyc2018.github.io/CS-Notes"> <img src="https://img.shields.io/badge/>-read-4ab8a1.svg"></a> <a href="https://xiaozhuanlan.com/CyC2018"> <img src="https://img.shields.io/badge/_-more-4ab8a1.svg"></a>
|
||||
<br> <br>
|
||||
本项目包含了技术面试必备的基础知识，浅显易懂，你不需要花很长的时间去阅读和理解成堆的技术书籍就可以快速掌握这些知识，从而节省宝贵的面试复习时间。你也可以订阅 <a href="https://xiaozhuanlan.com/CyC2018">面试进阶专栏</a>，包含了学习指导和面试技巧，让你更轻松拿到满意的 Offer。
|
||||
</div>
|
||||
|
||||
<br>
|
||||
|
||||
## :pencil2: Ⅰ 算法
|
||||
|
||||
- [剑指 Offer 题解](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/剑指%20offer%20题解.md)
|
||||
- [Leetcode 题解](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Leetcode%20题解.md)
|
||||
- [算法](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/算法.md)
|
||||
|
||||
## :computer: Ⅱ 操作系统
|
||||
|
||||
- [计算机操作系统](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/计算机操作系统.md)
|
||||
- [Linux](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Linux.md)
|
||||
|
||||
## :cloud: Ⅲ 网络
|
||||
|
||||
- [计算机网络](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/计算机网络.md)
|
||||
- [HTTP](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/HTTP.md)
|
||||
- [Socket](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Socket.md)
|
||||
|
||||
## :unlock: Ⅳ 面向对象
|
||||
|
||||
- [设计模式](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/设计模式.md)
|
||||
- [面向对象思想](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/面向对象思想.md)
|
||||
|
||||
## :floppy_disk: Ⅴ 数据库
|
||||
|
||||
- [数据库系统原理](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/数据库系统原理.md)
|
||||
- [SQL](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/SQL.md)
|
||||
- [Leetcode-Database 题解](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Leetcode-Database%20题解.md)
|
||||
- [MySQL](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/MySQL.md)
|
||||
- [Redis](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Redis.md)
|
||||
|
||||
## :coffee: Ⅵ Java
|
||||
|
||||
- [Java 基础](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Java%20基础.md)
|
||||
- [Java 容器](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Java%20容器.md)
|
||||
- [Java 并发](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Java%20并发.md)
|
||||
- [Java 虚拟机](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Java%20虚拟机.md)
|
||||
- [Java I/O](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Java%20IO.md)
|
||||
|
||||
## :bulb: Ⅶ 系统设计
|
||||
|
||||
- [系统设计基础](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/系统设计基础.md)
|
||||
- [分布式](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/分布式.md)
|
||||
- [集群](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/集群.md)
|
||||
- [攻击技术](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/攻击技术.md)
|
||||
- [缓存](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/缓存.md)
|
||||
- [消息队列](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/消息队列.md)
|
||||
|
||||
## :wrench: Ⅷ 工具
|
||||
|
||||
- [Git](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Git.md)
|
||||
- [Docker](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/Docker.md)
|
||||
- [构建工具](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/构建工具.md)
|
||||
- [正则表达式](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/正则表达式.md)
|
||||
|
||||
## :art: Ⅸ 编码实践
|
||||
|
||||
- [代码可读性](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/代码可读性.md)
|
||||
- [代码风格规范](https://github.com/CyC2018/CS-Notes/blob/master/docs/notes/代码风格规范.md)
|
||||
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## :memo: Ⅹ 后记
|
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### 内推信息
|
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||||
[Job-Recommend](https://github.com/CyC2018/Job-Recommend)
|
||||
|
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### 专栏
|
||||
|
||||
[面试进阶指南](https://xiaozhuanlan.com/CyC2018)
|
||||
|
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### 知识星球
|
||||
|
||||
想要向我提问关于学习和求职方面的建议？来知识星球，你的每个提问我都会认真回答。
|
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|
||||
[知识星球](other/Planet.md)
|
||||
|
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### QQ 群
|
||||
|
||||
为大家提供一个学习交流平台，在这里你可以自由地讨论技术问题。
|
||||
|
||||
<img src="https://github.com/CyC2018/CS-Notes/raw/master/other/group1.png" width="200px"></br>
|
||||
|
||||
### 排版
|
||||
|
||||
笔记内容按照 [文文案排版指北](https://github.com/sparanoid/chinese-copywriting-guidelines) 进行排版，以保证内容的可读性。笔记不使用 `![]()` 这种方式来引用图片，而是用 `<img>` 标签。一方面是为了能够控制图片以合适的大小显示，另一方面是因为 GFM 不支持 `<center> ![]() </center>` 让图片居中显示，只能使用 `<div align="center"> <img src=""/> </div>` 达到居中的效果。
|
||||
|
||||
我将自己实现的文档排版功能提取出来，放在 Github Page 中，无需下载安装即可免费使用：[Text-Typesetting](https://github.com/CyC2018/Text-Typesetting)。
|
||||
|
||||
### 上传方案
|
||||
|
||||
我在本地使用为知笔记软件进行书写，为了方便将本地笔记内容上传到 Github 上，实现了一整套自动化上传方案，包括文本文件的导出、提取图片、Markdown 文档转换、Git 同步。进行 Markdown 文档转换是因为 Github 使用的 GFM 不支持 MathJax 公式和 TOC 标记，所以需要替换 MathJax 公式为 CodeCogs 的云服务和重新生成 TOC 目录。
|
||||
|
||||
我将自己实现文档转换功能提取出来，方便大家在需要将本地 Markdown 上传到 Github，或者制作项目 README 文档时生成目录时使用：[GFM-Converter](https://github.com/CyC2018/GFM-Converter)。
|
||||
|
||||
### License
|
||||
|
||||
学习笔记不是从网上到处拼凑而来，除了少部分引用书上和技术文档的原文，其余都是笔者的原创。在您引用本仓库内容或者对内容进行修改演绎时，请署名并以相同方式共享，谢谢。
|
||||
|
||||
<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png" /></a>
|
||||
|
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|
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### Logo
|
||||
|
||||
Power by [logomakr](https://logomakr.com/).
|
||||
|
||||
### 致谢
|
||||
|
||||
感谢以下人员对本仓库做出的贡献，当然不仅仅只有这些贡献者，这里就不一一列举了。如果你希望被添加到这个名单中，并且提交过 Issue 或者 PR，请与笔者联系。
|
||||
|
||||
<a href="https://github.com/linw7">
|
||||
<img src="https://avatars3.githubusercontent.com/u/21679154?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/g10guang">
|
||||
<img src="https://avatars1.githubusercontent.com/u/18458140?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/ResolveWang">
|
||||
<img src="https://avatars1.githubusercontent.com/u/8018776?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/crossoverJie">
|
||||
<img src="https://avatars1.githubusercontent.com/u/15684156?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/jy03078584">
|
||||
<img src="https://avatars2.githubusercontent.com/u/7719370?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/kwongtailau">
|
||||
<img src="https://avatars0.githubusercontent.com/u/22954582?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/xiangflight">
|
||||
<img src="https://avatars2.githubusercontent.com/u/10072416?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/mafulong">
|
||||
<img src="https://avatars1.githubusercontent.com/u/24795000?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/yanglbme">
|
||||
<img src="https://avatars1.githubusercontent.com/u/21008209?s=400&v=4" width="50px">
|
||||
</a>
|
||||
|
||||
|
||||
|
||||
<!-- | Ⅰ | Ⅱ | Ⅲ | Ⅳ | Ⅴ | Ⅵ | Ⅶ | Ⅷ | Ⅸ | Ⅹ |
|
||||
| :--------: | :---------: | :---------: | :---------: | :---------: | :---------:| :---------: | :-------: | :-------:| :------:|
|
||||
| 算法[:pencil2:](#pencil2-算法) | 操作系统[:computer:](#computer-操作系统)|网络[:cloud:](#cloud-网络) | 面向对象[:couple:](#couple-面向对象) |数据库[:floppy_disk:](#floppy_disk-数据库)| Java [:coffee:](#coffee-java)| 系统设计[:bulb:](#bulb-系统设计)| 工具[:hammer:](#hammer-工具)| 编码实践[:speak_no_evil:](#speak_no_evil-编码实践)| 后记[:memo:](#memo-后记) | -->
|
||||
|
||||
<!-- | Ⅰ | Ⅱ | Ⅲ | Ⅳ | Ⅴ |
|
||||
| :--------: | :---------: | :---------: | :---------: | :---------: |
|
||||
|    算法   <br>[:pencil2:](#pencil2-算法) |    操作系统   <br>[:computer:](#computer-操作系统) |     网络    <br>[:cloud:](#cloud-网络) |    面向对象   <br>[:couple:](#couple-面向对象) |    数据库  <br>[:floppy_disk:](#floppy_disk-数据库)|
|
||||
|
||||
| Ⅵ | Ⅶ | Ⅷ | Ⅸ | Ⅹ |
|
||||
|:---------:| :---------: | :-------: | :-------:| :------:|
|
||||
|    Java   <br>[:coffee:](#coffee-java) |    系统设计   <br>[:bulb:](#bulb-系统设计) |     工具    <br>[:hammer:](#hammer-工具) |    编码实践   <br>[:speak_no_evil:](#speak_no_evil-编码实践) |    后记    <br>[:memo:](#memo-后记) |
|
||||
-->
|
||||
|
||||
<br>
|
||||
<div align="center">
|
||||
<a href="https://gitstar-ranking.com/repositories"> <img src="https://badgen.net/badge/Rank/20?icon=github&color=4ab8a1"></a>
|
||||
<a href="assets/download.md"> <img src="https://badgen.net/badge/OvO/%E7%A6%BB%E7%BA%BF%E4%B8%8B%E8%BD%BD?icon=telegram&color=4ab8a1"></a>
|
||||
<a href="https://cyc2018.github.io/CS-Notes"> <img src="https://badgen.net/badge/CyC/%E5%9C%A8%E7%BA%BF%E9%98%85%E8%AF%BB?icon=sourcegraph&color=4ab8a1"></a>
|
||||
<a href="#微信公众号"> <img src="https://badgen.net/badge/%e5%85%ac%e4%bc%97%e5%8f%b7/CyC2018?icon=rss&color=4ab8a1"></a>
|
||||
</div>
|
||||
<br>
|
||||
|
||||
| 算法 | 操作系统 | 网络 |面向对象| 数据库 | Java | 系统设计| 工具 |编码实践| 后记 |
|
||||
| :---: | :----: | :---: | :----: | :----: | :----: | :----: | :----: | :----: | :----: |
|
||||
| [:pencil2:](#pencil2-算法) | [:computer:](#computer-操作系统) | [:cloud:](#cloud-网络) | [:art:](#art-面向对象) | [:floppy_disk:](#floppy_disk-数据库) |[:coffee:](#coffee-java)| [:bulb:](#bulb-系统设计) |[:wrench:](#wrench-工具)| [:watermelon:](#watermelon-编码实践) |[:memo:](#memo-后记)|
|
||||
|
||||
<br>
|
||||
|
||||
<div align="center">
|
||||
<img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/LogoMakr_0zpEzN.png" width="200px">
|
||||
</div>
|
||||
|
||||
|
||||
|
||||
<br>
|
||||
|
||||
## :pencil2: 算法
|
||||
|
||||
- [剑指 Offer 题解](https://github.com/CyC2018/CS-Notes/blob/master/notes/剑指%20Offer%20题解%20-%20目录.md)
|
||||
- [Leetcode 题解](https://github.com/CyC2018/CS-Notes/blob/master/notes/Leetcode%20题解%20-%20目录.md)
|
||||
- [算法](https://github.com/CyC2018/CS-Notes/blob/master/notes/算法%20-%20目录.md)
|
||||
- [笔试面试题库](https://www.nowcoder.com/contestRoom?from=cyc_github)
|
||||
|
||||
## :computer: 操作系统
|
||||
|
||||
- [计算机操作系统](https://github.com/CyC2018/CS-Notes/blob/master/notes/计算机操作系统%20-%20目录.md)
|
||||
- [Linux](https://github.com/CyC2018/CS-Notes/blob/master/notes/Linux.md)
|
||||
|
||||
## :cloud: 网络
|
||||
|
||||
- [计算机网络](https://github.com/CyC2018/CS-Notes/blob/master/notes/计算机网络%20-%20目录.md)
|
||||
- [HTTP](https://github.com/CyC2018/CS-Notes/blob/master/notes/HTTP.md)
|
||||
- [Socket](https://github.com/CyC2018/CS-Notes/blob/master/notes/Socket.md)
|
||||
|
||||
## :art: 面向对象
|
||||
|
||||
- [面向对象思想](https://github.com/CyC2018/CS-Notes/blob/master/notes/面向对象思想.md)
|
||||
- [设计模式](https://github.com/CyC2018/CS-Notes/blob/master/notes/设计模式%20-%20目录.md)
|
||||
|
||||
## :floppy_disk: 数据库
|
||||
|
||||
- [数据库系统原理](https://github.com/CyC2018/CS-Notes/blob/master/notes/数据库系统原理.md)
|
||||
- [SQL](https://github.com/CyC2018/CS-Notes/blob/master/notes/SQL.md)
|
||||
- [Leetcode-Database 题解](https://github.com/CyC2018/CS-Notes/blob/master/notes/Leetcode-Database%20题解.md)
|
||||
- [MySQL](https://github.com/CyC2018/CS-Notes/blob/master/notes/MySQL.md)
|
||||
- [Redis](https://github.com/CyC2018/CS-Notes/blob/master/notes/Redis.md)
|
||||
|
||||
## :coffee: Java
|
||||
|
||||
- [Java 基础](https://github.com/CyC2018/CS-Notes/blob/master/notes/Java%20基础.md)
|
||||
- [Java 容器](https://github.com/CyC2018/CS-Notes/blob/master/notes/Java%20容器.md)
|
||||
- [Java 并发](https://github.com/CyC2018/CS-Notes/blob/master/notes/Java%20并发.md)
|
||||
- [Java 虚拟机](https://github.com/CyC2018/CS-Notes/blob/master/notes/Java%20虚拟机.md)
|
||||
- [Java I/O](https://github.com/CyC2018/CS-Notes/blob/master/notes/Java%20IO.md)
|
||||
|
||||
## :bulb: 系统设计
|
||||
|
||||
- [系统设计基础](https://github.com/CyC2018/CS-Notes/blob/master/notes/系统设计基础.md)
|
||||
- [分布式](https://github.com/CyC2018/CS-Notes/blob/master/notes/分布式.md)
|
||||
- [集群](https://github.com/CyC2018/CS-Notes/blob/master/notes/集群.md)
|
||||
- [攻击技术](https://github.com/CyC2018/CS-Notes/blob/master/notes/攻击技术.md)
|
||||
- [缓存](https://github.com/CyC2018/CS-Notes/blob/master/notes/缓存.md)
|
||||
- [消息队列](https://github.com/CyC2018/CS-Notes/blob/master/notes/消息队列.md)
|
||||
|
||||
## :wrench: 工具
|
||||
|
||||
- [Git](https://github.com/CyC2018/CS-Notes/blob/master/notes/Git.md)
|
||||
- [Docker](https://github.com/CyC2018/CS-Notes/blob/master/notes/Docker.md)
|
||||
- [构建工具](https://github.com/CyC2018/CS-Notes/blob/master/notes/构建工具.md)
|
||||
- [正则表达式](https://github.com/CyC2018/CS-Notes/blob/master/notes/正则表达式.md)
|
||||
|
||||
## :watermelon: 编码实践
|
||||
|
||||
- [代码可读性](https://github.com/CyC2018/CS-Notes/blob/master/notes/代码可读性.md)
|
||||
- [代码风格规范](https://github.com/CyC2018/CS-Notes/blob/master/notes/代码风格规范.md)
|
||||
|
||||
## :memo: 后记
|
||||
|
||||
<div align="center">
|
||||
<a href="https://www.nowcoder.com/discuss/137593?from=cyc_github"> 我的面经 </a> / <a href="https://cyc2018.github.io"> 我的简历 </a> / <a href="https://github.com/CyC2018/Markdown-Resume"> 简历模版 </a> / <a href="https://github.com/CyC2018/Job-Recommend"> 内推 </a> / <a href="https://xiaozhuanlan.com/CyC2018"> 专栏 </a> / <a href="assets/QQ2群.png"> QQ 群</a>
|
||||
<br><br>
|
||||
<img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img>
|
||||
</div>
|
||||
|
||||
|
||||
|
||||
### 排版
|
||||
|
||||
笔记内容按照 [中文文案排版指北](https://github.com/sparanoid/chinese-copywriting-guidelines/blob/master/README.zh-CN.md) 进行排版，以保证内容的可读性。
|
||||
|
||||
不使用 `![]()` 这种方式来引用图片，而是用 `<img>` 标签。一方面是为了能够控制图片以合适的大小显示，另一方面是因为 [GFM](https://github.github.com/gfm/) 不支持 `<center> ![]() </center>` 这种方法让图片居中显示，只能使用 `<div align="center"> <img src=""/> </div>` 达到居中的效果。
|
||||
|
||||
在线排版工具：[Text-Typesetting](https://github.com/CyC2018/Text-Typesetting)。
|
||||
|
||||
### License
|
||||
|
||||
本仓库的内容不是将网上的资料随意拼凑而来，除了少部分引用书上和技术文档的原文（这部分内容都在末尾的参考链接中加了出处），其余都是我的原创。在您引用本仓库内容或者对内容进行修改演绎时，请署名并以相同方式共享，谢谢。
|
||||
|
||||
转载文章请在开头明显处标明该页面地址，公众号等其它转载请联系 zhengyc101@163.com。
|
||||
|
||||
Logo：[logomakr](https://logomakr.com/)
|
||||
|
||||
<a rel="license" href="http://creativecommons.org/licenses/by-nc-sa/4.0/"><img alt="知识共享许可协议" style="border-width:0" src="https://i.creativecommons.org/l/by-nc-sa/4.0/88x31.png" /></a>
|
||||
|
||||
### 致谢
|
||||
|
||||
感谢以下人员对本仓库做出的贡献，当然不仅仅只有这些贡献者，这里就不一一列举了。如果你希望被添加到这个名单中，并且提交过 Issue 或者 PR，请与我联系。
|
||||
|
||||
<a href="https://github.com/linw7">
|
||||
<img src="https://avatars3.githubusercontent.com/u/21679154?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/g10guang">
|
||||
<img src="https://avatars1.githubusercontent.com/u/18458140?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/Sctwang">
|
||||
<img src="https://avatars3.githubusercontent.com/u/33345444?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/ResolveWang">
|
||||
<img src="https://avatars1.githubusercontent.com/u/8018776?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/crossoverJie">
|
||||
<img src="https://avatars1.githubusercontent.com/u/15684156?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/jy03078584">
|
||||
<img src="https://avatars2.githubusercontent.com/u/7719370?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/kwongtailau">
|
||||
<img src="https://avatars0.githubusercontent.com/u/22954582?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/xiangflight">
|
||||
<img src="https://avatars2.githubusercontent.com/u/10072416?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/mafulong">
|
||||
<img src="https://avatars1.githubusercontent.com/u/24795000?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/yanglbme">
|
||||
<img src="https://avatars1.githubusercontent.com/u/21008209?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/OOCZC">
|
||||
<img src="https://avatars1.githubusercontent.com/u/11623828?s=400&v=4" width="50px">
|
||||
</a>
|
||||
<a href="https://github.com/5renyuebing">
|
||||
<img src="https://avatars1.githubusercontent.com/u/32872430?s=400&v=4" width="50px">
|
||||
</a>
|
||||
|
||||
|
Before Width: | Height: | Size: 33 KiB After Width: | Height: | Size: 33 KiB |
BIN
assets/QQ2群.png
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BIN
assets/QQ2群.png
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Binary file not shown.
|
After Width: | Height: | Size: 56 KiB |
|
Before Width: | Height: | Size: 5.1 KiB After Width: | Height: | Size: 5.1 KiB |
BIN
assets/download-html.png
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BIN
assets/download-html.png
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|
After Width: | Height: | Size: 220 KiB |
BIN
assets/download-markdown.png
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BIN
assets/download-markdown.png
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Binary file not shown.
|
After Width: | Height: | Size: 271 KiB |
BIN
assets/download-pdf.png
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assets/download-pdf.png
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|
After Width: | Height: | Size: 220 KiB |
34
assets/download.md
Normal file
34
assets/download.md
Normal file
@ -0,0 +1,34 @@
|
||||
# 目的
|
||||
|
||||
考虑到有部分读者的网络环境较差，有时候在线访问速度很慢，导致阅读体验不佳。另外，PDF 等格式的离线版本相比于网页在线版本更方便做笔记。因此提供离线阅读版本给大家下载。
|
||||
|
||||
# 内容
|
||||
|
||||
有三种格式的离线版本：PDF、Markdown 和 HTML 。
|
||||
|
||||
## PDF
|
||||
|
||||
优点是方便做笔记；缺点是不能显示 GIF 图片（所以“剑指 Offer 题解”不建议使用 PDF 进行阅读），以及显示效果不佳。
|
||||
|
||||
|
||||
|
||||
## Markdown
|
||||
|
||||
优点是能很好地显示 GIF 图片，显示效果也很好；缺点是由于将所有内容整合在同一个文件中，导致实时渲染有点卡顿。
|
||||
|
||||
|
||||
|
||||
## HTML
|
||||
|
||||
优点是和 Markdown 的显示效果几乎一致，同时不需要 Markdown 的实时渲染，因此浏览速度更快；缺点是目录功能还不是很完善。
|
||||
|
||||
如果想在安卓手机端阅读，推荐使用这种格式，将 html 文件和图片文件都复制到手机上，用浏览器打开 html 文件并存成书签，以后就可以快速地离线阅读。
|
||||
|
||||
|
||||
|
||||
# 如何下载
|
||||
|
||||
离线版本由公众号 **CyC2018** 发布，最新版本也会在上面及时发布，在后台回复 **CyC** 即可获取下载链接。
|
||||
|
||||
<div align="center"><img width="350px" src="公众号二维码-1.png"></img></div>
|
||||
|
||||
BIN
assets/公众号二维码-2.png
Normal file
BIN
assets/公众号二维码-2.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 65 KiB |
BIN
assets/公众号海报7.png
Normal file
BIN
assets/公众号海报7.png
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 83 KiB |
119
docs/README.md
119
docs/README.md
@ -1,56 +1,63 @@
|
||||
|
||||
> [🍉 点击订阅面试进阶专栏](https://xiaozhuanlan.com/CyC2018)
|
||||
|
||||
## ✏️ 算法
|
||||
|
||||
> [剑指 Offer 题解](notes/剑指%20offer%20题解.md) </br>
|
||||
> [Leetcode 题解](notes/Leetcode%20题解) </br>
|
||||
> [算法](notes/算法.md)
|
||||
|
||||
## 💻 操作系统
|
||||
|
||||
> [计算机操作系统](notes/计算机操作系统.md) </br>
|
||||
> [Linux](notes/Linux.md)
|
||||
|
||||
## ☁️ 网络
|
||||
|
||||
> [计算机网络](notes/计算机网络.md) </br>
|
||||
> [HTTP](notes/HTTP.md) </br>
|
||||
> [Socket](notes/Socket.md)
|
||||
|
||||
## 🔓 面向对象
|
||||
|
||||
> [设计模式](notes/设计模式.md) </br>
|
||||
> [面向对象思想](notes/面向对象思想.md)
|
||||
|
||||
## 💾 数据库
|
||||
|
||||
> [数据库系统原理](notes/数据库系统原理.md) </br>
|
||||
> [SQL](notes/SQL.md) </br>
|
||||
> [Leetcode-Database 题解](notes/Leetcode-Database%20题解.md) </br>
|
||||
> [MySQL](notes/MySQL.md) </br>
|
||||
> [Redis](notes/Redis.md)
|
||||
|
||||
## ☕️ Java
|
||||
|
||||
> [Java 基础](notes/Java%20基础.md) </br>
|
||||
> [Java 容器](notes/Java%20容器.md) </br>
|
||||
> [Java 并发](notes/Java%20并发.md) </br>
|
||||
> [Java 虚拟机](notes/Java%20虚拟机.md) </br>
|
||||
> [Java I/O](notes/Java%20IO.md)
|
||||
|
||||
## 💡 系统设计
|
||||
|
||||
> [系统设计基础](notes/系统设计基础.md) </br>
|
||||
> [分布式](notes/分布式.md) </br>
|
||||
> [集群](notes/集群.md) </br>
|
||||
> [攻击技术](notes/攻击技术.md) </br>
|
||||
> [缓存](notes/缓存.md) </br>
|
||||
> [消息队列](notes/消息队列.md)
|
||||
|
||||
## 🔧 工具
|
||||
|
||||
> [Git](notes/Git.md) </br>
|
||||
> [Docker](notes/Docker.md) </br>
|
||||
> [正则表达式](notes/正则表达式.md) </br>
|
||||
> [构建工具](notes/构建工具.md)
|
||||
- [Github](https://github.com/CyC2018/CS-Notes)
|
||||
|
||||
## ✏️ 算法
|
||||
|
||||
- [剑指 Offer 题解](notes/剑指%20Offer%20题解%20-%20目录1.md) </br>
|
||||
- [Leetcode 题解](notes/Leetcode%20题解%20-%20目录1.md) </br>
|
||||
- [算法](notes/算法%20-%20目录1.md) </br>
|
||||
|
||||
## 💻 操作系统
|
||||
|
||||
- [计算机操作系统](notes/计算机操作系统%20-%20目录1.md) </br>
|
||||
- [Linux](notes/Linux.md)
|
||||
|
||||
## ☁️ 网络
|
||||
|
||||
- [计算机网络](notes/计算机网络%20-%20目录1.md) </br>
|
||||
- [HTTP](notes/HTTP.md) </br>
|
||||
- [Socket](notes/Socket.md)
|
||||
|
||||
## 🎨 面向对象
|
||||
|
||||
- [设计模式](notes/设计模式%20-%20目录1.md) </br>
|
||||
- [面向对象思想](notes/面向对象思想.md)
|
||||
|
||||
## 💾 数据库
|
||||
|
||||
- [数据库系统原理](notes/数据库系统原理.md) </br>
|
||||
- [SQL](notes/SQL.md) </br>
|
||||
- [Leetcode-Database 题解](notes/Leetcode-Database%20题解.md) </br>
|
||||
- [MySQL](notes/MySQL.md) </br>
|
||||
- [Redis](notes/Redis.md)
|
||||
|
||||
## ☕️ Java
|
||||
|
||||
- [Java 基础](notes/Java%20基础.md) </br>
|
||||
- [Java 容器](notes/Java%20容器.md) </br>
|
||||
- [Java 并发](notes/Java%20并发.md) </br>
|
||||
- [Java 虚拟机](notes/Java%20虚拟机.md) </br>
|
||||
- [Java I/O](notes/Java%20IO.md)
|
||||
|
||||
## 💡 系统设计
|
||||
|
||||
- [系统设计基础](notes/系统设计基础.md) </br>
|
||||
- [分布式](notes/分布式.md) </br>
|
||||
- [集群](notes/集群.md) </br>
|
||||
- [攻击技术](notes/攻击技术.md) </br>
|
||||
- [缓存](notes/缓存.md) </br>
|
||||
- [消息队列](notes/消息队列.md)
|
||||
|
||||
## 🔧 工具
|
||||
|
||||
- [Git](notes/Git.md) </br>
|
||||
- [Docker](notes/Docker.md) </br>
|
||||
- [正则表达式](notes/正则表达式.md) </br>
|
||||
- [构建工具](notes/构建工具.md)
|
||||
|
||||
<!--⭐️欢迎关注我的公众号 CyC2018，在公众号后台回复关键字 📚 **资料** 可领取复习大纲，这份大纲是我花了一整年时间整理的面试知识点列表，不仅系统整理了面试知识点，而且标注了各个知识点的重要程度，从而帮你理清多而杂的面试知识点。可以说我基本是按照这份大纲来进行复习的，这份大纲对我拿到了 BAT 头条等 Offer 起到很大的帮助。你们完全可以和我一样根据大纲上列的知识点来进行复习，就不用看很多不重要的内容，也可以知道哪些内容很重要从而多安排一些复习时间。
|
||||
<br/><br/>
|
||||
<div align="center">
|
||||
<img src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg" width="200px">
|
||||
</div> -->
|
||||
|
||||
|
||||
|
||||
@ -1,10 +1,11 @@
|
||||
<img width="150px" src="_media/LogoMakr_1J56bI.png">
|
||||
|
||||
# CS-Notes
|
||||
|
||||
- 本项目包含了技术面试必备的基础知识，浅显易懂，你不需要花很长的时间去阅读和理解成堆的技术书籍就可以快速掌握这些知识，从而节省宝贵的面试复习时间。你也可以订阅 <a href="https://xiaozhuanlan.com/CyC2018">面试进阶专栏</a>，包含了学习指导和面试技巧，让你更轻松拿到满意的 Offer。
|
||||
|
||||
<span id="busuanzi_container_site_pv">Site View : <span id="busuanzi_value_site_pv">
|
||||
|
||||
[Get Started](README.md)
|
||||
|
||||
<img width="220px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/other/LogoMakr_0zpEzN.png">
|
||||
|
||||
|
||||
- 本项目包含了技术面试必备的基础知识，内容浅显易懂，你不需要花很长的时间去阅读和理解成堆的技术书籍就可以快速掌握这些知识，从而节省宝贵的面试复习时间。
|
||||
|
||||
<!--<span id="busuanzi_container_site_pv">Site View : <span id="busuanzi_value_site_pv">-->
|
||||
|
||||
[](https://github.com/CyC2018/CS-Notes) [](https://github.com/CyC2018/CS-Notes)
|
||||
|
||||
[开始阅读](README.md)
|
||||
|
||||
|
||||
BIN
docs/_media/公众号.jpg
Normal file
BIN
docs/_media/公众号.jpg
Normal file
Binary file not shown.
|
After Width: | Height: | Size: 27 KiB |
2050
docs/_style/bootstrap-4.0.0-dist/css/bootstrap-grid.css
vendored
2050
docs/_style/bootstrap-4.0.0-dist/css/bootstrap-grid.css
vendored
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Load Diff
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@ -1,330 +0,0 @@
|
||||
/*!
|
||||
* Bootstrap Reboot v4.0.0 (https://getbootstrap.com)
|
||||
* Copyright 2011-2018 The Bootstrap Authors
|
||||
* Copyright 2011-2018 Twitter, Inc.
|
||||
* Licensed under MIT (https://github.com/twbs/bootstrap/blob/master/LICENSE)
|
||||
* Forked from Normalize.css, licensed MIT (https://github.com/necolas/normalize.css/blob/master/LICENSE.md)
|
||||
*/
|
||||
*,
|
||||
*::before,
|
||||
*::after {
|
||||
box-sizing: border-box;
|
||||
}
|
||||
|
||||
html {
|
||||
font-family: sans-serif;
|
||||
line-height: 1.15;
|
||||
-webkit-text-size-adjust: 100%;
|
||||
-ms-text-size-adjust: 100%;
|
||||
-ms-overflow-style: scrollbar;
|
||||
-webkit-tap-highlight-color: transparent;
|
||||
}
|
||||
|
||||
@-ms-viewport {
|
||||
width: device-width;
|
||||
}
|
||||
|
||||
article, aside, dialog, figcaption, figure, footer, header, hgroup, main, nav, section {
|
||||
display: block;
|
||||
}
|
||||
|
||||
body {
|
||||
margin: 0;
|
||||
font-family: -apple-system, BlinkMacSystemFont, "Segoe UI", Roboto, "Helvetica Neue", Arial, sans-serif, "Apple Color Emoji", "Segoe UI Emoji", "Segoe UI Symbol";
|
||||
font-size: 1rem;
|
||||
font-weight: 400;
|
||||
line-height: 1.5;
|
||||
color: #212529;
|
||||
text-align: left;
|
||||
background-color: #fff;
|
||||
}
|
||||
|
||||
[tabindex="-1"]:focus {
|
||||
outline: 0 !important;
|
||||
}
|
||||
|
||||
hr {
|
||||
box-sizing: content-box;
|
||||
height: 0;
|
||||
overflow: visible;
|
||||
}
|
||||
|
||||
h1, h2, h3, h4, h5, h6 {
|
||||
margin-top: 0;
|
||||
margin-bottom: 0.5rem;
|
||||
}
|
||||
|
||||
p {
|
||||
margin-top: 0;
|
||||
margin-bottom: 1rem;
|
||||
}
|
||||
|
||||
abbr[title],
|
||||
abbr[data-original-title] {
|
||||
text-decoration: underline;
|
||||
-webkit-text-decoration: underline dotted;
|
||||
text-decoration: underline dotted;
|
||||
cursor: help;
|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
dl {
|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
ol ul,
|
||||
ul ol {
|
||||
margin-bottom: 0;
|
||||
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|
||||
|
||||
dt {
|
||||
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|
||||
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|
||||
|
||||
dd {
|
||||
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|
||||
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|
||||
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|
||||
|
||||
blockquote {
|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
|
||||
sub,
|
||||
sup {
|
||||
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|
||||
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|
||||
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|
||||
vertical-align: baseline;
|
||||
}
|
||||
|
||||
sub {
|
||||
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|
||||
}
|
||||
|
||||
sup {
|
||||
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|
||||
}
|
||||
|
||||
a {
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
pre {
|
||||
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|
||||
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|
||||
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|
||||
-ms-overflow-style: scrollbar;
|
||||
}
|
||||
|
||||
figure {
|
||||
margin: 0 0 1rem;
|
||||
}
|
||||
|
||||
img {
|
||||
vertical-align: middle;
|
||||
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|
||||
}
|
||||
|
||||
svg:not(:root) {
|
||||
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|
||||
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|
||||
|
||||
table {
|
||||
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|
||||
}
|
||||
|
||||
caption {
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
button {
|
||||
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|
||||
}
|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
optgroup,
|
||||
textarea {
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
[type="submit"] {
|
||||
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|
||||
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|
||||
|
||||
button::-moz-focus-inner,
|
||||
[type="button"]::-moz-focus-inner,
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
input[type="radio"],
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
fieldset {
|
||||
min-width: 0;
|
||||
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|
||||
margin: 0;
|
||||
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|
||||
}
|
||||
|
||||
legend {
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
progress {
|
||||
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|
||||
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|
||||
|
||||
[type="number"]::-webkit-inner-spin-button,
|
||||
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|
||||
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|
||||
}
|
||||
|
||||
[type="search"] {
|
||||
outline-offset: -2px;
|
||||
-webkit-appearance: none;
|
||||
}
|
||||
|
||||
[type="search"]::-webkit-search-cancel-button,
|
||||
[type="search"]::-webkit-search-decoration {
|
||||
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|
||||
}
|
||||
|
||||
::-webkit-file-upload-button {
|
||||
font: inherit;
|
||||
-webkit-appearance: button;
|
||||
}
|
||||
|
||||
output {
|
||||
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|
||||
}
|
||||
|
||||
summary {
|
||||
display: list-item;
|
||||
cursor: pointer;
|
||||
}
|
||||
|
||||
template {
|
||||
display: none;
|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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@ -1,8 +0,0 @@
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/*!
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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@ -1,330 +0,0 @@
|
||||
/*!
|
||||
* Bootstrap Reboot v4.0.0 (https://getbootstrap.com)
|
||||
* Copyright 2011-2018 The Bootstrap Authors
|
||||
* Copyright 2011-2018 Twitter, Inc.
|
||||
* Licensed under MIT (https://github.com/twbs/bootstrap/blob/master/LICENSE)
|
||||
* Forked from Normalize.css, licensed MIT (https://github.com/necolas/normalize.css/blob/master/LICENSE.md)
|
||||
*/
|
||||
*,
|
||||
*::before,
|
||||
*::after {
|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
line-height: 1.15;
|
||||
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|
||||
-ms-text-size-adjust: 100%;
|
||||
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|
||||
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|
||||
}
|
||||
|
||||
@-ms-viewport {
|
||||
width: device-width;
|
||||
}
|
||||
|
||||
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|
||||
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|
||||
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|
||||
|
||||
body {
|
||||
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|
||||
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|
||||
font-size: 1rem;
|
||||
font-weight: 400;
|
||||
line-height: 1.5;
|
||||
color: #212529;
|
||||
text-align: left;
|
||||
background-color: #fff;
|
||||
}
|
||||
|
||||
[tabindex="-1"]:focus {
|
||||
outline: 0 !important;
|
||||
}
|
||||
|
||||
hr {
|
||||
box-sizing: content-box;
|
||||
height: 0;
|
||||
overflow: visible;
|
||||
}
|
||||
|
||||
h1, h2, h3, h4, h5, h6 {
|
||||
margin-top: 0;
|
||||
margin-bottom: 0.5rem;
|
||||
}
|
||||
|
||||
p {
|
||||
margin-top: 0;
|
||||
margin-bottom: 1rem;
|
||||
}
|
||||
|
||||
abbr[title],
|
||||
abbr[data-original-title] {
|
||||
text-decoration: underline;
|
||||
-webkit-text-decoration: underline dotted;
|
||||
text-decoration: underline dotted;
|
||||
cursor: help;
|
||||
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|
||||
}
|
||||
|
||||
address {
|
||||
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|
||||
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|
||||
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|
||||
}
|
||||
|
||||
ol,
|
||||
ul,
|
||||
dl {
|
||||
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|
||||
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|
||||
}
|
||||
|
||||
ol ol,
|
||||
ul ul,
|
||||
ol ul,
|
||||
ul ol {
|
||||
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|
||||
}
|
||||
|
||||
dt {
|
||||
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|
||||
}
|
||||
|
||||
dd {
|
||||
margin-bottom: .5rem;
|
||||
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|
||||
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|
||||
|
||||
blockquote {
|
||||
margin: 0 0 1rem;
|
||||
}
|
||||
|
||||
dfn {
|
||||
font-style: italic;
|
||||
}
|
||||
|
||||
b,
|
||||
strong {
|
||||
font-weight: bolder;
|
||||
}
|
||||
|
||||
small {
|
||||
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|
||||
}
|
||||
|
||||
sub,
|
||||
sup {
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
}
|
||||
|
||||
sub {
|
||||
bottom: -.25em;
|
||||
}
|
||||
|
||||
sup {
|
||||
top: -.5em;
|
||||
}
|
||||
|
||||
a {
|
||||
color: #007bff;
|
||||
text-decoration: none;
|
||||
background-color: transparent;
|
||||
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|
||||
}
|
||||
|
||||
a:hover {
|
||||
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|
||||
text-decoration: underline;
|
||||
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|
||||
|
||||
a:not([href]):not([tabindex]) {
|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
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|
||||
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|
||||
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|
||||
|
||||
pre,
|
||||
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|
||||
kbd,
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
|
||||
pre {
|
||||
margin-top: 0;
|
||||
margin-bottom: 1rem;
|
||||
overflow: auto;
|
||||
-ms-overflow-style: scrollbar;
|
||||
}
|
||||
|
||||
figure {
|
||||
margin: 0 0 1rem;
|
||||
}
|
||||
|
||||
img {
|
||||
vertical-align: middle;
|
||||
border-style: none;
|
||||
}
|
||||
|
||||
svg:not(:root) {
|
||||
overflow: hidden;
|
||||
}
|
||||
|
||||
table {
|
||||
border-collapse: collapse;
|
||||
}
|
||||
|
||||
caption {
|
||||
padding-top: 0.75rem;
|
||||
padding-bottom: 0.75rem;
|
||||
color: #6c757d;
|
||||
text-align: left;
|
||||
caption-side: bottom;
|
||||
}
|
||||
|
||||
th {
|
||||
text-align: inherit;
|
||||
}
|
||||
|
||||
label {
|
||||
display: inline-block;
|
||||
margin-bottom: .5rem;
|
||||
}
|
||||
|
||||
button {
|
||||
border-radius: 0;
|
||||
}
|
||||
|
||||
button:focus {
|
||||
outline: 1px dotted;
|
||||
outline: 5px auto -webkit-focus-ring-color;
|
||||
}
|
||||
|
||||
input,
|
||||
button,
|
||||
select,
|
||||
optgroup,
|
||||
textarea {
|
||||
margin: 0;
|
||||
font-family: inherit;
|
||||
font-size: inherit;
|
||||
line-height: inherit;
|
||||
}
|
||||
|
||||
button,
|
||||
input {
|
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overflow: visible;
|
||||
}
|
||||
|
||||
button,
|
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select {
|
||||
text-transform: none;
|
||||
}
|
||||
|
||||
button,
|
||||
html [type="button"],
|
||||
[type="reset"],
|
||||
[type="submit"] {
|
||||
-webkit-appearance: button;
|
||||
}
|
||||
|
||||
button::-moz-focus-inner,
|
||||
[type="button"]::-moz-focus-inner,
|
||||
[type="reset"]::-moz-focus-inner,
|
||||
[type="submit"]::-moz-focus-inner {
|
||||
padding: 0;
|
||||
border-style: none;
|
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}
|
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|
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input[type="radio"],
|
||||
input[type="checkbox"] {
|
||||
box-sizing: border-box;
|
||||
padding: 0;
|
||||
}
|
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|
||||
input[type="date"],
|
||||
input[type="time"],
|
||||
input[type="datetime-local"],
|
||||
input[type="month"] {
|
||||
-webkit-appearance: listbox;
|
||||
}
|
||||
|
||||
textarea {
|
||||
overflow: auto;
|
||||
resize: vertical;
|
||||
}
|
||||
|
||||
fieldset {
|
||||
min-width: 0;
|
||||
padding: 0;
|
||||
margin: 0;
|
||||
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|
||||
}
|
||||
|
||||
legend {
|
||||
display: block;
|
||||
width: 100%;
|
||||
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|
||||
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|
||||
margin-bottom: .5rem;
|
||||
font-size: 1.5rem;
|
||||
line-height: inherit;
|
||||
color: inherit;
|
||||
white-space: normal;
|
||||
}
|
||||
|
||||
progress {
|
||||
vertical-align: baseline;
|
||||
}
|
||||
|
||||
[type="number"]::-webkit-inner-spin-button,
|
||||
[type="number"]::-webkit-outer-spin-button {
|
||||
height: auto;
|
||||
}
|
||||
|
||||
[type="search"] {
|
||||
outline-offset: -2px;
|
||||
-webkit-appearance: none;
|
||||
}
|
||||
|
||||
[type="search"]::-webkit-search-cancel-button,
|
||||
[type="search"]::-webkit-search-decoration {
|
||||
-webkit-appearance: none;
|
||||
}
|
||||
|
||||
::-webkit-file-upload-button {
|
||||
font: inherit;
|
||||
-webkit-appearance: button;
|
||||
}
|
||||
|
||||
output {
|
||||
display: inline-block;
|
||||
}
|
||||
|
||||
summary {
|
||||
display: list-item;
|
||||
cursor: pointer;
|
||||
}
|
||||
|
||||
template {
|
||||
display: none;
|
||||
}
|
||||
|
||||
[hidden] {
|
||||
display: none !important;
|
||||
}
|
||||
/*# sourceMappingURL=bootstrap-reboot.css.map */
|
||||
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@ -1,8 +0,0 @@
|
||||
/*!
|
||||
* Bootstrap Reboot v4.0.0 (https://getbootstrap.com)
|
||||
* Copyright 2011-2018 The Bootstrap Authors
|
||||
* Copyright 2011-2018 Twitter, Inc.
|
||||
* Licensed under MIT (https://github.com/twbs/bootstrap/blob/master/LICENSE)
|
||||
* Forked from Normalize.css, licensed MIT (https://github.com/necolas/normalize.css/blob/master/LICENSE.md)
|
||||
*/*,::after,::before{box-sizing:border-box}html{font-family:sans-serif;line-height:1.15;-webkit-text-size-adjust:100%;-ms-text-size-adjust:100%;-ms-overflow-style:scrollbar;-webkit-tap-highlight-color:transparent}@-ms-viewport{width:device-width}article,aside,dialog,figcaption,figure,footer,header,hgroup,main,nav,section{display:block}body{margin:0;font-family:-apple-system,BlinkMacSystemFont,"Segoe UI",Roboto,"Helvetica Neue",Arial,sans-serif,"Apple Color Emoji","Segoe UI Emoji","Segoe UI Symbol";font-size:1rem;font-weight:400;line-height:1.5;color:#212529;text-align:left;background-color:#fff}[tabindex="-1"]:focus{outline:0!important}hr{box-sizing:content-box;height:0;overflow:visible}h1,h2,h3,h4,h5,h6{margin-top:0;margin-bottom:.5rem}p{margin-top:0;margin-bottom:1rem}abbr[data-original-title],abbr[title]{text-decoration:underline;-webkit-text-decoration:underline dotted;text-decoration:underline dotted;cursor:help;border-bottom:0}address{margin-bottom:1rem;font-style:normal;line-height:inherit}dl,ol,ul{margin-top:0;margin-bottom:1rem}ol ol,ol ul,ul ol,ul ul{margin-bottom:0}dt{font-weight:700}dd{margin-bottom:.5rem;margin-left:0}blockquote{margin:0 0 1rem}dfn{font-style:italic}b,strong{font-weight:bolder}small{font-size:80%}sub,sup{position:relative;font-size:75%;line-height:0;vertical-align:baseline}sub{bottom:-.25em}sup{top:-.5em}a{color:#007bff;text-decoration:none;background-color:transparent;-webkit-text-decoration-skip:objects}a:hover{color:#0056b3;text-decoration:underline}a:not([href]):not([tabindex]){color:inherit;text-decoration:none}a:not([href]):not([tabindex]):focus,a:not([href]):not([tabindex]):hover{color:inherit;text-decoration:none}a:not([href]):not([tabindex]):focus{outline:0}code,kbd,pre,samp{font-family:monospace,monospace;font-size:1em}pre{margin-top:0;margin-bottom:1rem;overflow:auto;-ms-overflow-style:scrollbar}figure{margin:0 0 1rem}img{vertical-align:middle;border-style:none}svg:not(:root){overflow:hidden}table{border-collapse:collapse}caption{padding-top:.75rem;padding-bottom:.75rem;color:#6c757d;text-align:left;caption-side:bottom}th{text-align:inherit}label{display:inline-block;margin-bottom:.5rem}button{border-radius:0}button:focus{outline:1px dotted;outline:5px auto -webkit-focus-ring-color}button,input,optgroup,select,textarea{margin:0;font-family:inherit;font-size:inherit;line-height:inherit}button,input{overflow:visible}button,select{text-transform:none}[type=reset],[type=submit],button,html [type=button]{-webkit-appearance:button}[type=button]::-moz-focus-inner,[type=reset]::-moz-focus-inner,[type=submit]::-moz-focus-inner,button::-moz-focus-inner{padding:0;border-style:none}input[type=checkbox],input[type=radio]{box-sizing:border-box;padding:0}input[type=date],input[type=datetime-local],input[type=month],input[type=time]{-webkit-appearance:listbox}textarea{overflow:auto;resize:vertical}fieldset{min-width:0;padding:0;margin:0;border:0}legend{display:block;width:100%;max-width:100%;padding:0;margin-bottom:.5rem;font-size:1.5rem;line-height:inherit;color:inherit;white-space:normal}progress{vertical-align:baseline}[type=number]::-webkit-inner-spin-button,[type=number]::-webkit-outer-spin-button{height:auto}[type=search]{outline-offset:-2px;-webkit-appearance:none}[type=search]::-webkit-search-cancel-button,[type=search]::-webkit-search-decoration{-webkit-appearance:none}::-webkit-file-upload-button{font:inherit;-webkit-appearance:button}output{display:inline-block}summary{display:list-item;cursor:pointer}template{display:none}[hidden]{display:none!important}
|
||||
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|
||||
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@ -1,51 +0,0 @@
|
||||
/*隐藏头部的目录*/
|
||||
#main>ul:nth-child(1) {
|
||||
display: none;
|
||||
}
|
||||
|
||||
#main>ul:nth-child(2) {
|
||||
display: none;
|
||||
}
|
||||
|
||||
.markdown-section h1 {
|
||||
margin: 3rem 0 2rem 0;
|
||||
}
|
||||
|
||||
.markdown-section h2 {
|
||||
margin: 2rem 0 1rem;
|
||||
}
|
||||
|
||||
.content,
|
||||
.sidebar,
|
||||
.markdown-section,
|
||||
body,
|
||||
.search input,
|
||||
.sidebar-toggle {
|
||||
background-color: rgba(243, 242, 238, 1) !important;
|
||||
}
|
||||
|
||||
body {
|
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/*font-family: Microsoft YaHei, Source Sans Pro, Helvetica Neue, Arial, sans-serif !important;*/
|
||||
}
|
||||
|
||||
.markdown-section>p {
|
||||
font-size: 16px !important;
|
||||
}
|
||||
|
||||
.markdown-section pre>code {
|
||||
font-family: Consolas, Roboto Mono, Monaco, courier, monospace !important;
|
||||
font-size: .9rem !important;
|
||||
|
||||
}
|
||||
|
||||
/*.anchor span {
|
||||
color: rgb(66, 185, 131);
|
||||
}*/
|
||||
|
||||
section.cover h1 {
|
||||
margin: 0;
|
||||
}
|
||||
|
||||
body>section>div.cover-main>ul>li>a {
|
||||
color: #42b983;
|
||||
}
|
||||
504
docs/index.html
504
docs/index.html
@ -1,67 +1,439 @@
|
||||
<!DOCTYPE html>
|
||||
<html lang="en">
|
||||
|
||||
<head>
|
||||
<meta charset="UTF-8">
|
||||
<title>CS-Notes</title>
|
||||
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1" />
|
||||
<meta name="description" content="Description">
|
||||
<meta name="viewport" content="width=device-width, user-scalable=no, initial-scale=1.0, maximum-scale=1.0, minimum-scale=1.0">
|
||||
<link rel="icon" href="_media/logomakr-3sxxzw-128x128.png">
|
||||
<link rel="stylesheet" href="//unpkg.com/docsify/lib/themes/vue.css">
|
||||
<link rel="stylesheet" href="//unpkg.com/gitalk/dist/gitalk.css">
|
||||
<link rel="stylesheet" href="_style/style.css">
|
||||
<!--solarizedlight tomorrow coy-->
|
||||
<link rel="stylesheet" href="_style/prism-master/themes/prism-coy.css">
|
||||
</head>
|
||||
|
||||
<body>
|
||||
<!-- <nav>
|
||||
<a href="#/README">HOME</a>
|
||||
</nav> -->
|
||||
<div id="app"></div>
|
||||
<script>
|
||||
window.$docsify = {
|
||||
maxAge: 100,
|
||||
name: 'CS-Notes',
|
||||
repo: 'https://github.com/CyC2018/CS-Notes',
|
||||
ga: 'UA-121566133-2',
|
||||
search: {
|
||||
paths: 'auto',
|
||||
placeholder: '🔍 Type to search ',
|
||||
noData: '😞 No Results! ',
|
||||
// Headline depth, 1 - 6
|
||||
depth: 6
|
||||
},
|
||||
// subMaxLevel: 2,
|
||||
coverpage: true
|
||||
}
|
||||
</script>
|
||||
<script src="//unpkg.com/docsify/lib/docsify.min.js"></script>
|
||||
<!-- <script src="//unpkg.com/docsify/lib/plugins/search.min.js"></script> -->
|
||||
<!-- 上面的基本不可用，无法搜索 -->
|
||||
<script src="https://cdn.bootcss.com/docsify/4.5.9/plugins/search.min.js"></script>
|
||||
<script src="//unpkg.com/docsify/lib/plugins/gitalk.min.js"></script>
|
||||
<script src="//unpkg.com/gitalk/dist/gitalk.min.js"></script>
|
||||
<script src="//unpkg.com/docsify-copy-code"></script>
|
||||
<script src="//unpkg.com/prismjs/components/prism-java.min.js"></script>
|
||||
<script src="//unpkg.com/prismjs/components/prism-c.min.js"></script>
|
||||
<script src="//unpkg.com/prismjs/components/prism-bash.min.js"></script>
|
||||
<script src="//unpkg.com/prismjs/components/prism-sql.min.js"></script>
|
||||
<script src="//unpkg.com/docsify/lib/plugins/zoom-image.min.js"></script>
|
||||
<!-- <script src="//unpkg.com/docsify/lib/plugins/emoji.min.js"></script> -->
|
||||
<script src="//unpkg.com/docsify/lib/plugins/ga.min.js"></script>
|
||||
<script async src="//busuanzi.ibruce.info/busuanzi/2.3/busuanzi.pure.mini.js"></script>
|
||||
<script>
|
||||
const gitalk = new Gitalk({
|
||||
clientID: 'c6c02367e36ca6e2bb2d',
|
||||
clientSecret: '31a2700e3315b21c5e9f2e887709d8cf21f9ff8f',
|
||||
repo: 'Gittalk-Issue',
|
||||
owner: 'CyC2018',
|
||||
admin: ['CyC2018'],
|
||||
// distractionFreeMode: false
|
||||
})
|
||||
</script>
|
||||
</body>
|
||||
|
||||
<!DOCTYPE html>
|
||||
<html lang="en">
|
||||
|
||||
<head>
|
||||
<meta charset="UTF-8">
|
||||
<title>CS-Notes</title>
|
||||
<meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1" />
|
||||
<meta name="description" content="Description">
|
||||
<meta name="viewport" content="width=device-width, user-scalable=yes, initial-scale=1.0, maximum-scale=2.0, minimum-scale=1.0">
|
||||
<link rel="icon" href=" https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/LogoMakr_1J56bI.png">
|
||||
<link rel="stylesheet" href="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/vue.css">
|
||||
|
||||
<!-- 将自定义样式放在 Github 上会导致加载速度变得非常慢，所以采取直接内嵌的方式 -->
|
||||
<style type="text/css">
|
||||
|
||||
/* 隐藏头部的目录 */
|
||||
#main>ul:nth-child(1) {
|
||||
display: none;
|
||||
}
|
||||
|
||||
#main>ul:nth-child(2) {
|
||||
display: none;
|
||||
}
|
||||
|
||||
h1+ul {
|
||||
display: block !important;
|
||||
}
|
||||
|
||||
.markdown-section h1 {
|
||||
margin: 3rem 0 2rem 0;
|
||||
}
|
||||
|
||||
.markdown-section h2 {
|
||||
margin: 2rem 0 1rem;
|
||||
}
|
||||
|
||||
img,
|
||||
pre {
|
||||
border-radius: 5px;
|
||||
}
|
||||
|
||||
.markdown-section p.tip,
|
||||
.markdown-section tr:nth-child(1n) {
|
||||
background-color: #f8f8f8 !important;
|
||||
}
|
||||
|
||||
.content,
|
||||
.sidebar,
|
||||
.markdown-section,
|
||||
body,
|
||||
.search input {
|
||||
background-color: rgba(243, 242, 238, 1) !important;
|
||||
}
|
||||
|
||||
@media (min-width:600px) {
|
||||
.sidebar-toggle {
|
||||
background-color: #f3f2ee;
|
||||
}
|
||||
}
|
||||
|
||||
.docsify-copy-code-button {
|
||||
background: #f8f8f8 !important;
|
||||
color: #7a7a7a !important;
|
||||
}
|
||||
|
||||
body {
|
||||
/*font-family: Microsoft YaHei, Source Sans Pro, Helvetica Neue, Arial, sans-serif !important;*/
|
||||
}
|
||||
|
||||
.markdown-section pre>code {
|
||||
font-size: 13px;
|
||||
}
|
||||
|
||||
code,
|
||||
pre {
|
||||
background-color: #fff !important;
|
||||
}
|
||||
|
||||
.markdown-section>p {
|
||||
font-size: 16px !important;
|
||||
}
|
||||
|
||||
.markdown-section pre>code {
|
||||
font-family: Consolas, Roboto Mono, Monaco, courier, monospace !important;
|
||||
}
|
||||
|
||||
p, h1, h2, h3, h4, ol, ul {
|
||||
letter-spacing: 2px !important;
|
||||
}
|
||||
|
||||
p, ol, ul {
|
||||
line-height: 30px !important;
|
||||
}
|
||||
|
||||
@media (min-width:600px) {
|
||||
.markdown-section pre>code {
|
||||
font-size: .9rem !important;
|
||||
letter-spacing: 1.1px !important;
|
||||
}
|
||||
}
|
||||
|
||||
@media (max-width:600px) {
|
||||
.markdown-section pre>code {
|
||||
padding-top: 5px;
|
||||
padding-bottom: 5px;
|
||||
padding-left: 15px !important;
|
||||
}
|
||||
|
||||
pre:after {
|
||||
content: "" !important;
|
||||
}
|
||||
}
|
||||
|
||||
|
||||
/*.anchor span {
|
||||
color: rgb(66, 185, 131);
|
||||
}*/
|
||||
|
||||
section.cover h1 {
|
||||
margin: 0;
|
||||
}
|
||||
|
||||
body>section>div.cover-main>ul>li>a {
|
||||
color: #42b983;
|
||||
}
|
||||
|
||||
.markdown-section > div > img,
|
||||
.markdown-section pre {
|
||||
box-shadow: 0px 0px 20px 11px #eaeaea;
|
||||
}
|
||||
|
||||
pre {
|
||||
background-color: #f3f2ee !important;
|
||||
}
|
||||
|
||||
@media (min-width:600px) {
|
||||
pre code {
|
||||
/*box-shadow: 2px 1px 20px 2px #aaa;*/
|
||||
/*border-radius: 10px !important;*/
|
||||
padding-left: 20px !important;
|
||||
}
|
||||
}
|
||||
|
||||
@media (max-width:600px) {
|
||||
pre {
|
||||
padding-left: 3px !important;
|
||||
padding-right: 3px !important;
|
||||
margin-left: -20px !important;
|
||||
margin-right: -20px !important;
|
||||
box-shadow: 0px 0px 20px 0px #eee !important;
|
||||
}
|
||||
|
||||
|
||||
.docsify-copy-code-button {
|
||||
display: none;
|
||||
}
|
||||
}
|
||||
|
||||
.markdown-section pre {
|
||||
padding-left: 0 !important;
|
||||
padding-right: 0px !important;
|
||||
}
|
||||
</style>
|
||||
<style type="text/css">
|
||||
/**
|
||||
* prism.js Coy theme for JavaScript, CoffeeScript, CSS and HTML
|
||||
* Based on https://github.com/tshedor/workshop-wp-theme (Example: http://workshop.kansan.com/category/sessions/basics or http://workshop.timshedor.com/category/sessions/basics);
|
||||
* @author Tim Shedor
|
||||
*/
|
||||
|
||||
code[class*="language-"],
|
||||
pre[class*="language-"] {
|
||||
color: black;
|
||||
background: none;
|
||||
font-family: Consolas, Monaco, 'Andale Mono', 'Ubuntu Mono', monospace;
|
||||
text-align: left;
|
||||
white-space: pre;
|
||||
word-spacing: normal;
|
||||
word-break: normal;
|
||||
word-wrap: normal;
|
||||
line-height: 1.5;
|
||||
|
||||
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|
||||
-o-tab-size: 4;
|
||||
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|
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|
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|
||||
-moz-hyphens: none;
|
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|
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|
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}
|
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|
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/* Code blocks */
|
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pre[class*="language-"] {
|
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position: relative;
|
||||
margin: .5em 0;
|
||||
overflow: visible;
|
||||
padding: 0;
|
||||
}
|
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|
||||
pre[class*="language-"]>code {
|
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position: relative;
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|
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box-shadow: -1px 0px 0px 0px #358ccb, 0px 0px 0px 1px #dfdfdf;
|
||||
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|
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|
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|
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|
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|
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|
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|
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|
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height: inherit;
|
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|
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|
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/* Margin bottom to accommodate shadow */
|
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|
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|
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|
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|
||||
margin-bottom: 1em;
|
||||
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|
||||
|
||||
/* Inline code */
|
||||
:not(pre)>code[class*="language-"] {
|
||||
position: relative;
|
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|
||||
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|
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|
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|
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|
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|
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pre[class*="language-"]:before,
|
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|
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|
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|
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|
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
box-shadow: 0px 13px 8px #979797;
|
||||
-webkit-transform: rotate(-2deg);
|
||||
-moz-transform: rotate(-2deg);
|
||||
-ms-transform: rotate(-2deg);
|
||||
-o-transform: rotate(-2deg);
|
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transform: rotate(-2deg);
|
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}
|
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|
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:not(pre)>code[class*="language-"]:after,
|
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pre[class*="language-"]:after {
|
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right: 0.75em;
|
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left: auto;
|
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-webkit-transform: rotate(2deg);
|
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-moz-transform: rotate(2deg);
|
||||
-ms-transform: rotate(2deg);
|
||||
-o-transform: rotate(2deg);
|
||||
transform: rotate(2deg);
|
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}
|
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|
||||
/*黑色*/
|
||||
.token.function,
|
||||
.token.operator,
|
||||
.token.class-name {
|
||||
color: #2C3E50;
|
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}
|
||||
|
||||
/*深蓝加粗*/
|
||||
.token.keyword {
|
||||
color: #333;
|
||||
font-weight: 700;
|
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}
|
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|
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|
||||
/*浅蓝*/
|
||||
.token.property,
|
||||
.token.tag,
|
||||
.token.boolean,
|
||||
.token.number,
|
||||
.token.function-name,
|
||||
.token.constant,
|
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.token.symbol,
|
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.token.deleted {
|
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color: #2980B9;
|
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}
|
||||
|
||||
.token.comment,
|
||||
.token.block-comment,
|
||||
.token.prolog,
|
||||
.token.doctype,
|
||||
.token.cdata {
|
||||
color: #7D8B99;
|
||||
}
|
||||
|
||||
.token.punctuation {
|
||||
color: #5F6364;
|
||||
}
|
||||
|
||||
|
||||
.token.selector,
|
||||
.token.attr-name,
|
||||
.token.string,
|
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.token.char,
|
||||
.token.builtin,
|
||||
.token.inserted {
|
||||
color: #1ABC9C;
|
||||
font-weight: bold;
|
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}
|
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|
||||
.token.entity,
|
||||
.token.url,
|
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.token.variable {
|
||||
color: #a67f59;
|
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/*background: rgba(255, 255, 255, 0.5);*/
|
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}
|
||||
|
||||
.token.atrule,
|
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.token.attr-value {
|
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color: #1990b8;
|
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}
|
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|
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.token.regex,
|
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.token.important {
|
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color: #e90;
|
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}
|
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|
||||
.language-css .token.string,
|
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.style .token.string {
|
||||
color: #a67f59;
|
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background: rgba(255, 255, 255, 0.5);
|
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}
|
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|
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.token.important {
|
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font-weight: normal;
|
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}
|
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|
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.token.bold {
|
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font-weight: bold;
|
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}
|
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|
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.token.italic {
|
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font-style: italic;
|
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}
|
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|
||||
.token.entity {
|
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cursor: help;
|
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}
|
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|
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.namespace {
|
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opacity: .7;
|
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}
|
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|
||||
@media screen and (max-width: 767px) {
|
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|
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pre[class*="language-"]:before,
|
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pre[class*="language-"]:after {
|
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bottom: 14px;
|
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box-shadow: none;
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}
|
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|
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}
|
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|
||||
/* Plugin styles */
|
||||
.token.tab:not(:empty):before,
|
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.token.cr:before,
|
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.token.lf:before {
|
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color: #e0d7d1;
|
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}
|
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|
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/* Plugin styles: Line Numbers */
|
||||
pre[class*="language-"].line-numbers.line-numbers {
|
||||
padding-left: 0;
|
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}
|
||||
|
||||
pre[class*="language-"].line-numbers.line-numbers code {
|
||||
padding-left: 3.8em;
|
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}
|
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|
||||
pre[class*="language-"].line-numbers.line-numbers .line-numbers-rows {
|
||||
left: 0;
|
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}
|
||||
|
||||
/* Plugin styles: Line Highlight */
|
||||
pre[class*="language-"][data-line] {
|
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padding-top: 0;
|
||||
padding-bottom: 0;
|
||||
padding-left: 0;
|
||||
}
|
||||
|
||||
pre[data-line] code {
|
||||
position: relative;
|
||||
padding-left: 4em;
|
||||
}
|
||||
|
||||
pre .line-highlight {
|
||||
margin-top: 0;
|
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}
|
||||
</style>
|
||||
</head>
|
||||
|
||||
<body>
|
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<div id="app"></div>
|
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<script>
|
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window.$docsify = {
|
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maxAge: 100,
|
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name: 'CS-Notes',
|
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repo: 'https://github.com/CyC2018/CS-Notes',
|
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search: {
|
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paths: 'auto',
|
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placeholder: '🔍 Type to search ',
|
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noData: '😞 No Results! ',
|
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depth: 6
|
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},
|
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// subMaxLevel: 2,
|
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coverpage: true
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}
|
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</script>
|
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<script src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/docsify.min.js"></script>
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<script src="https://cdn.bootcss.com/docsify/4.5.9/plugins/search.min.js"></script>
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<script src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/docsify-copy-code.min.js"></script>
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<script src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/prism-java.min.js"></script>
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<script src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/prism-c.min.js"></script>
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<script src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/prism-bash.min.js"></script>
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<script src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/prism-sql.min.js"></script>
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<script src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/zoom-image.min.js"></script>
|
||||
<!-- <script async src="//busuanzi.ibruce.info/busuanzi/2.3/busuanzi.pure.mini.js"></script> -->
|
||||
</body>
|
||||
|
||||
</html>
|
||||
76
docs/notes/10.1 斐波那契数列.md
Normal file
76
docs/notes/10.1 斐波那契数列.md
Normal file
@ -0,0 +1,76 @@
|
||||
# 10.1 斐波那契数列
|
||||
|
||||
## 题目链接
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/c6c7742f5ba7442aada113136ddea0c3?tpId=13&tqId=11160&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
求斐波那契数列的第 n 项，n <= 39。
|
||||
|
||||
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?f(n)=\left\{\begin{array}{rcl}0&&{n=0}\\1&&{n=1}\\f(n-1)+f(n-2)&&{n>1}\end{array}\right." class="mathjax-pic"/></div> <br> -->
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/45be9587-6069-4ab7-b9ac-840db1a53744.jpg" width="330px"> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
如果使用递归求解，会重复计算一些子问题。例如，计算 f(4) 需要计算 f(3) 和 f(2)，计算 f(3) 需要计算 f(2) 和 f(1)，可以看到 f(2) 被重复计算了。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/c13e2a3d-b01c-4a08-a69b-db2c4e821e09.png" width="350px"/> </div><br>
|
||||
|
||||
递归是将一个问题划分成多个子问题求解，动态规划也是如此，但是动态规划会把子问题的解缓存起来，从而避免重复求解子问题。
|
||||
|
||||
```java
|
||||
public int Fibonacci(int n) {
|
||||
if (n <= 1)
|
||||
return n;
|
||||
int[] fib = new int[n + 1];
|
||||
fib[1] = 1;
|
||||
for (int i = 2; i <= n; i++)
|
||||
fib[i] = fib[i - 1] + fib[i - 2];
|
||||
return fib[n];
|
||||
}
|
||||
```
|
||||
|
||||
考虑到第 i 项只与第 i-1 和第 i-2 项有关，因此只需要存储前两项的值就能求解第 i 项，从而将空间复杂度由 O(N) 降低为 O(1)。
|
||||
|
||||
```java
|
||||
public int Fibonacci(int n) {
|
||||
if (n <= 1)
|
||||
return n;
|
||||
int pre2 = 0, pre1 = 1;
|
||||
int fib = 0;
|
||||
for (int i = 2; i <= n; i++) {
|
||||
fib = pre2 + pre1;
|
||||
pre2 = pre1;
|
||||
pre1 = fib;
|
||||
}
|
||||
return fib;
|
||||
}
|
||||
```
|
||||
|
||||
由于待求解的 n 小于 40，因此可以将前 40 项的结果先进行计算，之后就能以 O(1) 时间复杂度得到第 n 项的值。
|
||||
|
||||
```java
|
||||
public class Solution {
|
||||
|
||||
private int[] fib = new int[40];
|
||||
|
||||
public Solution() {
|
||||
fib[1] = 1;
|
||||
for (int i = 2; i < fib.length; i++)
|
||||
fib[i] = fib[i - 1] + fib[i - 2];
|
||||
}
|
||||
|
||||
public int Fibonacci(int n) {
|
||||
return fib[n];
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
49
docs/notes/10.2 矩形覆盖.md
Normal file
49
docs/notes/10.2 矩形覆盖.md
Normal file
@ -0,0 +1,49 @@
|
||||
# 10.2 矩形覆盖
|
||||
|
||||
## 题目链接
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/72a5a919508a4251859fb2cfb987a0e6?tpId=13&tqId=11163&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
我们可以用 2\*1 的小矩形横着或者竖着去覆盖更大的矩形。请问用 n 个 2\*1 的小矩形无重叠地覆盖一个 2\*n 的大矩形，总共有多少种方法？
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/b903fda8-07d0-46a7-91a7-e803892895cf.gif" width="100px"> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
当 n 为 1 时，只有一种覆盖方法：
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/f6e146f1-57ad-411b-beb3-770a142164ef.png" width="100px"> </div><br>
|
||||
|
||||
当 n 为 2 时，有两种覆盖方法：
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/fb3b8f7a-4293-4a38-aae1-62284db979a3.png" width="200px"> </div><br>
|
||||
|
||||
要覆盖 2\*n 的大矩形，可以先覆盖 2\*1 的矩形，再覆盖 2\*(n-1) 的矩形；或者先覆盖 2\*2 的矩形，再覆盖 2\*(n-2) 的矩形。而覆盖 2\*(n-1) 和 2\*(n-2) 的矩形可以看成子问题。该问题的递推公式如下：
|
||||
|
||||
<!-- <div align="center"><img src="https://latex.codecogs.com/gif.latex?f(n)=\left\{\begin{array}{rcl}1&&{n=1}\\2&&{n=2}\\f(n-1)+f(n-2)&&{n>1}\end{array}\right." class="mathjax-pic"/></div> <br> -->
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/508c6e52-9f93-44ed-b6b9-e69050e14807.jpg" width="370px"> </div><br>
|
||||
|
||||
```java
|
||||
public int RectCover(int n) {
|
||||
if (n <= 2)
|
||||
return n;
|
||||
int pre2 = 1, pre1 = 2;
|
||||
int result = 0;
|
||||
for (int i = 3; i <= n; i++) {
|
||||
result = pre2 + pre1;
|
||||
pre2 = pre1;
|
||||
pre1 = result;
|
||||
}
|
||||
return result;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
47
docs/notes/10.3 跳台阶.md
Normal file
47
docs/notes/10.3 跳台阶.md
Normal file
@ -0,0 +1,47 @@
|
||||
# 10.3 跳台阶
|
||||
|
||||
## 题目链接
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/8c82a5b80378478f9484d87d1c5f12a4?tpId=13&tqId=11161&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
一只青蛙一次可以跳上 1 级台阶，也可以跳上 2 级。求该青蛙跳上一个 n 级的台阶总共有多少种跳法。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/9dae7475-934f-42e5-b3b3-12724337170a.png" width="380px"> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
当 n = 1 时，只有一种跳法：
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/72aac98a-d5df-4bfa-a71a-4bb16a87474c.png" width="250px"> </div><br>
|
||||
|
||||
当 n = 2 时，有两种跳法：
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/1b80288d-1b35-4cd3-aa17-7e27ab9a2389.png" width="300px"> </div><br>
|
||||
|
||||
跳 n 阶台阶，可以先跳 1 阶台阶，再跳 n-1 阶台阶；或者先跳 2 阶台阶，再跳 n-2 阶台阶。而 n-1 和 n-2 阶台阶的跳法可以看成子问题，该问题的递推公式为：
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/508c6e52-9f93-44ed-b6b9-e69050e14807.jpg" width="350px"> </div><br>
|
||||
|
||||
```java
|
||||
public int JumpFloor(int n) {
|
||||
if (n <= 2)
|
||||
return n;
|
||||
int pre2 = 1, pre1 = 2;
|
||||
int result = 0;
|
||||
for (int i = 2; i < n; i++) {
|
||||
result = pre2 + pre1;
|
||||
pre2 = pre1;
|
||||
pre1 = result;
|
||||
}
|
||||
return result;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
67
docs/notes/10.4 变态跳台阶.md
Normal file
67
docs/notes/10.4 变态跳台阶.md
Normal file
@ -0,0 +1,67 @@
|
||||
# 10.4 变态跳台阶
|
||||
|
||||
## 题目链接
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/22243d016f6b47f2a6928b4313c85387?tpId=13&tqId=11162&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
一只青蛙一次可以跳上 1 级台阶，也可以跳上 2 级... 它也可以跳上 n 级。求该青蛙跳上一个 n 级的台阶总共有多少种跳法。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/cd411a94-3786-4c94-9e08-f28320e010d5.png" width="380px"> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
### 动态规划
|
||||
|
||||
```java
|
||||
public int JumpFloorII(int target) {
|
||||
int[] dp = new int[target];
|
||||
Arrays.fill(dp, 1);
|
||||
for (int i = 1; i < target; i++)
|
||||
for (int j = 0; j < i; j++)
|
||||
dp[i] += dp[j];
|
||||
return dp[target - 1];
|
||||
}
|
||||
```
|
||||
|
||||
### 数学推导
|
||||
|
||||
跳上 n-1 级台阶，可以从 n-2 级跳 1 级上去，也可以从 n-3 级跳 2 级上去...，那么
|
||||
|
||||
```
|
||||
f(n-1) = f(n-2) + f(n-3) + ... + f(0)
|
||||
```
|
||||
|
||||
同样，跳上 n 级台阶，可以从 n-1 级跳 1 级上去，也可以从 n-2 级跳 2 级上去... ，那么
|
||||
|
||||
```
|
||||
f(n) = f(n-1) + f(n-2) + ... + f(0)
|
||||
```
|
||||
|
||||
综上可得
|
||||
|
||||
```
|
||||
f(n) - f(n-1) = f(n-1)
|
||||
```
|
||||
|
||||
即
|
||||
|
||||
```
|
||||
f(n) = 2*f(n-1)
|
||||
```
|
||||
|
||||
所以 f(n) 是一个等比数列
|
||||
|
||||
```source-java
|
||||
public int JumpFloorII(int target) {
|
||||
return (int) Math.pow(2, target - 1);
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
72
docs/notes/11. 旋转数组的最小数字.md
Normal file
72
docs/notes/11. 旋转数组的最小数字.md
Normal file
@ -0,0 +1,72 @@
|
||||
# 11. 旋转数组的最小数字
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/9f3231a991af4f55b95579b44b7a01ba?tpId=13&tqId=11159&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
把一个数组最开始的若干个元素搬到数组的末尾，我们称之为数组的旋转。输入一个非递减排序的数组的一个旋转，输出旋转数组的最小元素。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/0038204c-4b8a-42a5-921d-080f6674f989.png" width="210px"> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
将旋转数组对半分可以得到一个包含最小元素的新旋转数组，以及一个非递减排序的数组。新的旋转数组的数组元素是原数组的一半，从而将问题规模减少了一半，这种折半性质的算法的时间复杂度为 O(logN)（为了方便，这里将 log<sub>2</sub>N 写为 logN）。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/424f34ab-a9fd-49a6-9969-d76b42251365.png" width="300px"> </div><br>
|
||||
|
||||
此时问题的关键在于确定对半分得到的两个数组哪一个是旋转数组，哪一个是非递减数组。我们很容易知道非递减数组的第一个元素一定小于等于最后一个元素。
|
||||
|
||||
通过修改二分查找算法进行求解（l 代表 low，m 代表 mid，h 代表 high）：
|
||||
|
||||
- 当 nums[m] <= nums[h] 时，表示 [m, h] 区间内的数组是非递减数组，[l, m] 区间内的数组是旋转数组，此时令 h = m；
|
||||
- 否则 [m + 1, h] 区间内的数组是旋转数组，令 l = m + 1。
|
||||
|
||||
```java
|
||||
public int minNumberInRotateArray(int[] nums) {
|
||||
if (nums.length == 0)
|
||||
return 0;
|
||||
int l = 0, h = nums.length - 1;
|
||||
while (l < h) {
|
||||
int m = l + (h - l) / 2;
|
||||
if (nums[m] <= nums[h])
|
||||
h = m;
|
||||
else
|
||||
l = m + 1;
|
||||
}
|
||||
return nums[l];
|
||||
}
|
||||
```
|
||||
|
||||
如果数组元素允许重复，会出现一个特殊的情况：nums[l] == nums[m] == nums[h]，此时无法确定解在哪个区间，需要切换到顺序查找。例如对于数组 {1,1,1,0,1}，l、m 和 h 指向的数都为 1，此时无法知道最小数字 0 在哪个区间。
|
||||
|
||||
```java
|
||||
public int minNumberInRotateArray(int[] nums) {
|
||||
if (nums.length == 0)
|
||||
return 0;
|
||||
int l = 0, h = nums.length - 1;
|
||||
while (l < h) {
|
||||
int m = l + (h - l) / 2;
|
||||
if (nums[l] == nums[m] && nums[m] == nums[h])
|
||||
return minNumber(nums, l, h);
|
||||
else if (nums[m] <= nums[h])
|
||||
h = m;
|
||||
else
|
||||
l = m + 1;
|
||||
}
|
||||
return nums[l];
|
||||
}
|
||||
|
||||
private int minNumber(int[] nums, int l, int h) {
|
||||
for (int i = l; i < h; i++)
|
||||
if (nums[i] > nums[i + 1])
|
||||
return nums[i + 1];
|
||||
return nums[l];
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
71
docs/notes/12. 矩阵中的路径.md
Normal file
71
docs/notes/12. 矩阵中的路径.md
Normal file
@ -0,0 +1,71 @@
|
||||
# 12. 矩阵中的路径
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/c61c6999eecb4b8f88a98f66b273a3cc?tpId=13&tqId=11218&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
判断在一个矩阵中是否存在一条包含某字符串所有字符的路径。路径可以从矩阵中的任意一个格子开始，每一步可以在矩阵中向上下左右移动一个格子。如果一条路径经过了矩阵中的某一个格子，则该路径不能再进入该格子。
|
||||
|
||||
例如下面的矩阵包含了一条 bfce 路径。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/1db1c7ea-0443-478b-8df9-7e33b1336cc4.png" width="200px"> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
使用回溯法（backtracking）进行求解，它是一种暴力搜索方法，通过搜索所有可能的结果来求解问题。回溯法在一次搜索结束时需要进行回溯（回退），将这一次搜索过程中设置的状态进行清除，从而开始一次新的搜索过程。例如下图示例中，从 f 开始，下一步有 4 种搜索可能，如果先搜索 b，需要将 b 标记为已经使用，防止重复使用。在这一次搜索结束之后，需要将 b 的已经使用状态清除，并搜索 c。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/dc964b86-7a08-4bde-a3d9-e6ddceb29f98.png" width="200px"> </div><br>
|
||||
|
||||
本题的输入是数组而不是矩阵（二维数组），因此需要先将数组转换成矩阵。
|
||||
|
||||
```java
|
||||
private final static int[][] next = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
|
||||
private int rows;
|
||||
private int cols;
|
||||
|
||||
public boolean hasPath(char[] array, int rows, int cols, char[] str) {
|
||||
if (rows == 0 || cols == 0) return false;
|
||||
this.rows = rows;
|
||||
this.cols = cols;
|
||||
boolean[][] marked = new boolean[rows][cols];
|
||||
char[][] matrix = buildMatrix(array);
|
||||
for (int i = 0; i < rows; i++)
|
||||
for (int j = 0; j < cols; j++)
|
||||
if (backtracking(matrix, str, marked, 0, i, j))
|
||||
return true;
|
||||
|
||||
return false;
|
||||
}
|
||||
|
||||
private boolean backtracking(char[][] matrix, char[] str,
|
||||
boolean[][] marked, int pathLen, int r, int c) {
|
||||
|
||||
if (pathLen == str.length) return true;
|
||||
if (r < 0 || r >= rows || c < 0 || c >= cols
|
||||
|| matrix[r][c] != str[pathLen] || marked[r][c]) {
|
||||
|
||||
return false;
|
||||
}
|
||||
marked[r][c] = true;
|
||||
for (int[] n : next)
|
||||
if (backtracking(matrix, str, marked, pathLen + 1, r + n[0], c + n[1]))
|
||||
return true;
|
||||
marked[r][c] = false;
|
||||
return false;
|
||||
}
|
||||
|
||||
private char[][] buildMatrix(char[] array) {
|
||||
char[][] matrix = new char[rows][cols];
|
||||
for (int r = 0, idx = 0; r < rows; r++)
|
||||
for (int c = 0; c < cols; c++)
|
||||
matrix[r][c] = array[idx++];
|
||||
return matrix;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
65
docs/notes/13. 机器人的运动范围.md
Normal file
65
docs/notes/13. 机器人的运动范围.md
Normal file
@ -0,0 +1,65 @@
|
||||
# 13. 机器人的运动范围
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/6e5207314b5241fb83f2329e89fdecc8?tpId=13&tqId=11219&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
地上有一个 m 行和 n 列的方格。一个机器人从坐标 (0, 0) 的格子开始移动，每一次只能向左右上下四个方向移动一格，但是不能进入行坐标和列坐标的数位之和大于 k 的格子。
|
||||
|
||||
例如，当 k 为 18 时，机器人能够进入方格 (35,37)，因为 3+5+3+7=18。但是，它不能进入方格 (35,38)，因为 3+5+3+8=19。请问该机器人能够达到多少个格子？
|
||||
|
||||
## 解题思路
|
||||
|
||||
使用深度优先搜索（Depth First Search，DFS）方法进行求解。回溯是深度优先搜索的一种特例，它在一次搜索过程中需要设置一些本次搜索过程的局部状态，并在本次搜索结束之后清除状态。而普通的深度优先搜索并不需要使用这些局部状态，虽然还是有可能设置一些全局状态。
|
||||
|
||||
```java
|
||||
private static final int[][] next = {{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
|
||||
private int cnt = 0;
|
||||
private int rows;
|
||||
private int cols;
|
||||
private int threshold;
|
||||
private int[][] digitSum;
|
||||
|
||||
public int movingCount(int threshold, int rows, int cols) {
|
||||
this.rows = rows;
|
||||
this.cols = cols;
|
||||
this.threshold = threshold;
|
||||
initDigitSum();
|
||||
boolean[][] marked = new boolean[rows][cols];
|
||||
dfs(marked, 0, 0);
|
||||
return cnt;
|
||||
}
|
||||
|
||||
private void dfs(boolean[][] marked, int r, int c) {
|
||||
if (r < 0 || r >= rows || c < 0 || c >= cols || marked[r][c])
|
||||
return;
|
||||
marked[r][c] = true;
|
||||
if (this.digitSum[r][c] > this.threshold)
|
||||
return;
|
||||
cnt++;
|
||||
for (int[] n : next)
|
||||
dfs(marked, r + n[0], c + n[1]);
|
||||
}
|
||||
|
||||
private void initDigitSum() {
|
||||
int[] digitSumOne = new int[Math.max(rows, cols)];
|
||||
for (int i = 0; i < digitSumOne.length; i++) {
|
||||
int n = i;
|
||||
while (n > 0) {
|
||||
digitSumOne[i] += n % 10;
|
||||
n /= 10;
|
||||
}
|
||||
}
|
||||
this.digitSum = new int[rows][cols];
|
||||
for (int i = 0; i < this.rows; i++)
|
||||
for (int j = 0; j < this.cols; j++)
|
||||
this.digitSum[i][j] = digitSumOne[i] + digitSumOne[j];
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
59
docs/notes/14. 剪绳子.md
Normal file
59
docs/notes/14. 剪绳子.md
Normal file
@ -0,0 +1,59 @@
|
||||
# 14. 剪绳子
|
||||
|
||||
[Leetcode](https://leetcode.com/problems/integer-break/description/)
|
||||
|
||||
## 题目描述
|
||||
|
||||
把一根绳子剪成多段，并且使得每段的长度乘积最大。
|
||||
|
||||
```html
|
||||
n = 2
|
||||
return 1 (2 = 1 + 1)
|
||||
|
||||
n = 10
|
||||
return 36 (10 = 3 + 3 + 4)
|
||||
```
|
||||
|
||||
## 解题思路
|
||||
|
||||
### 贪心
|
||||
|
||||
尽可能多剪长度为 3 的绳子，并且不允许有长度为 1 的绳子出现。如果出现了，就从已经切好长度为 3 的绳子中拿出一段与长度为 1 的绳子重新组合，把它们切成两段长度为 2 的绳子。
|
||||
|
||||
证明：当 n >= 5 时，3(n - 3) - n = 2n - 9 > 0，且 2(n - 2) - n = n - 4 > 0。因此在 n >= 5 的情况下，将绳子剪成一段为 2 或者 3，得到的乘积会更大。又因为 3(n - 3) - 2(n - 2) = n - 5 >= 0，所以剪成一段长度为 3 比长度为 2 得到的乘积更大。
|
||||
|
||||
```java
|
||||
public int integerBreak(int n) {
|
||||
if (n < 2)
|
||||
return 0;
|
||||
if (n == 2)
|
||||
return 1;
|
||||
if (n == 3)
|
||||
return 2;
|
||||
int timesOf3 = n / 3;
|
||||
if (n - timesOf3 * 3 == 1)
|
||||
timesOf3--;
|
||||
int timesOf2 = (n - timesOf3 * 3) / 2;
|
||||
return (int) (Math.pow(3, timesOf3)) * (int) (Math.pow(2, timesOf2));
|
||||
}
|
||||
```
|
||||
|
||||
### 动态规划
|
||||
|
||||
```java
|
||||
public int integerBreak(int n) {
|
||||
int[] dp = new int[n + 1];
|
||||
dp[1] = 1;
|
||||
for (int i = 2; i <= n; i++)
|
||||
for (int j = 1; j < i; j++)
|
||||
dp[i] = Math.max(dp[i], Math.max(j * (i - j), dp[j] * (i - j)));
|
||||
return dp[n];
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
47
docs/notes/15. 二进制中 1 的个数.md
Normal file
47
docs/notes/15. 二进制中 1 的个数.md
Normal file
@ -0,0 +1,47 @@
|
||||
# 15. 二进制中 1 的个数
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/8ee967e43c2c4ec193b040ea7fbb10b8?tpId=13&tqId=11164&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
输入一个整数，输出该数二进制表示中 1 的个数。
|
||||
|
||||
### n&(n-1)
|
||||
|
||||
该位运算去除 n 的位级表示中最低的那一位。
|
||||
|
||||
```
|
||||
n : 10110100
|
||||
n-1 : 10110011
|
||||
n&(n-1) : 10110000
|
||||
```
|
||||
|
||||
时间复杂度：O(M)，其中 M 表示 1 的个数。
|
||||
|
||||
|
||||
```java
|
||||
public int NumberOf1(int n) {
|
||||
int cnt = 0;
|
||||
while (n != 0) {
|
||||
cnt++;
|
||||
n &= (n - 1);
|
||||
}
|
||||
return cnt;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
### Integer.bitCount()
|
||||
|
||||
```java
|
||||
public int NumberOf1(int n) {
|
||||
return Integer.bitCount(n);
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
43
docs/notes/16. 数值的整数次方.md
Normal file
43
docs/notes/16. 数值的整数次方.md
Normal file
@ -0,0 +1,43 @@
|
||||
# 16. 数值的整数次方
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/1a834e5e3e1a4b7ba251417554e07c00?tpId=13&tqId=11165&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
给定一个 double 类型的浮点数 base 和 int 类型的整数 exponent，求 base 的 exponent 次方。
|
||||
|
||||
## 解题思路
|
||||
|
||||
下面的讨论中 x 代表 base，n 代表 exponent。
|
||||
|
||||
<!--<div align="center"><img src="https://latex.codecogs.com/gif.latex?x^n=\left\{\begin{array}{rcl}(x*x)^{n/2}&&{n\%2=0}\\x*(x*x)^{n/2}&&{n\%2=1}\end{array}\right." class="mathjax-pic"/></div> <br>-->
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/48b1d459-8832-4e92-938a-728aae730739.jpg" width="330px"> </div><br>
|
||||
|
||||
|
||||
因为 (x\*x)<sup>n/2</sup> 可以通过递归求解，并且每次递归 n 都减小一半，因此整个算法的时间复杂度为 O(logN)。
|
||||
|
||||
```java
|
||||
public double Power(double base, int exponent) {
|
||||
if (exponent == 0)
|
||||
return 1;
|
||||
if (exponent == 1)
|
||||
return base;
|
||||
boolean isNegative = false;
|
||||
if (exponent < 0) {
|
||||
exponent = -exponent;
|
||||
isNegative = true;
|
||||
}
|
||||
double pow = Power(base * base, exponent / 2);
|
||||
if (exponent % 2 != 0)
|
||||
pow = pow * base;
|
||||
return isNegative ? 1 / pow : pow;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
47
docs/notes/17. 打印从 1 到最大的 n 位数.md
Normal file
47
docs/notes/17. 打印从 1 到最大的 n 位数.md
Normal file
@ -0,0 +1,47 @@
|
||||
# 17. 打印从 1 到最大的 n 位数
|
||||
|
||||
## 题目描述
|
||||
|
||||
输入数字 n，按顺序打印出从 1 到最大的 n 位十进制数。比如输入 3，则打印出 1、2、3 一直到最大的 3 位数即 999。
|
||||
|
||||
## 解题思路
|
||||
|
||||
由于 n 可能会非常大，因此不能直接用 int 表示数字，而是用 char 数组进行存储。
|
||||
|
||||
使用回溯法得到所有的数。
|
||||
|
||||
```java
|
||||
public void print1ToMaxOfNDigits(int n) {
|
||||
if (n <= 0)
|
||||
return;
|
||||
char[] number = new char[n];
|
||||
print1ToMaxOfNDigits(number, 0);
|
||||
}
|
||||
|
||||
private void print1ToMaxOfNDigits(char[] number, int digit) {
|
||||
if (digit == number.length) {
|
||||
printNumber(number);
|
||||
return;
|
||||
}
|
||||
for (int i = 0; i < 10; i++) {
|
||||
number[digit] = (char) (i + '0');
|
||||
print1ToMaxOfNDigits(number, digit + 1);
|
||||
}
|
||||
}
|
||||
|
||||
private void printNumber(char[] number) {
|
||||
int index = 0;
|
||||
while (index < number.length && number[index] == '0')
|
||||
index++;
|
||||
while (index < number.length)
|
||||
System.out.print(number[index++]);
|
||||
System.out.println();
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
44
docs/notes/18.1 在 O(1) 时间内删除链表节点.md
Normal file
44
docs/notes/18.1 在 O(1) 时间内删除链表节点.md
Normal file
@ -0,0 +1,44 @@
|
||||
# 18.1 在 O(1) 时间内删除链表节点
|
||||
|
||||
## 解题思路
|
||||
|
||||
① 如果该节点不是尾节点，那么可以直接将下一个节点的值赋给该节点，然后令该节点指向下下个节点，再删除下一个节点，时间复杂度为 O(1)。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/1176f9e1-3442-4808-a47a-76fbaea1b806.png" width="600"/> </div><br>
|
||||
|
||||
② 否则，就需要先遍历链表，找到节点的前一个节点，然后让前一个节点指向 null，时间复杂度为 O(N)。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/4bf8d0ba-36f0-459e-83a0-f15278a5a157.png" width="600"/> </div><br>
|
||||
|
||||
综上，如果进行 N 次操作，那么大约需要操作节点的次数为 N-1+N=2N-1，其中 N-1 表示 N-1 个不是尾节点的每个节点以 O(1) 的时间复杂度操作节点的总次数，N 表示 1 个尾节点以 O(N) 的时间复杂度操作节点的总次数。(2N-1)/N \~ 2，因此该算法的平均时间复杂度为 O(1)。
|
||||
|
||||
```java
|
||||
public ListNode deleteNode(ListNode head, ListNode tobeDelete) {
|
||||
if (head == null || tobeDelete == null)
|
||||
return null;
|
||||
if (tobeDelete.next != null) {
|
||||
// 要删除的节点不是尾节点
|
||||
ListNode next = tobeDelete.next;
|
||||
tobeDelete.val = next.val;
|
||||
tobeDelete.next = next.next;
|
||||
} else {
|
||||
if (head == tobeDelete)
|
||||
// 只有一个节点
|
||||
head = null;
|
||||
else {
|
||||
ListNode cur = head;
|
||||
while (cur.next != tobeDelete)
|
||||
cur = cur.next;
|
||||
cur.next = null;
|
||||
}
|
||||
}
|
||||
return head;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
32
docs/notes/18.2 删除链表中重复的结点.md
Normal file
32
docs/notes/18.2 删除链表中重复的结点.md
Normal file
@ -0,0 +1,32 @@
|
||||
# 18.2 删除链表中重复的结点
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/fc533c45b73a41b0b44ccba763f866ef?tpId=13&tqId=11209&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/17e301df-52e8-4886-b593-841a16d13e44.png" width="450"/> </div><br>
|
||||
|
||||
## 解题描述
|
||||
|
||||
```java
|
||||
public ListNode deleteDuplication(ListNode pHead) {
|
||||
if (pHead == null || pHead.next == null)
|
||||
return pHead;
|
||||
ListNode next = pHead.next;
|
||||
if (pHead.val == next.val) {
|
||||
while (next != null && pHead.val == next.val)
|
||||
next = next.next;
|
||||
return deleteDuplication(next);
|
||||
} else {
|
||||
pHead.next = deleteDuplication(pHead.next);
|
||||
return pHead;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
47
docs/notes/19. 正则表达式匹配.md
Normal file
47
docs/notes/19. 正则表达式匹配.md
Normal file
@ -0,0 +1,47 @@
|
||||
# 19. 正则表达式匹配
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/45327ae22b7b413ea21df13ee7d6429c?tpId=13&tqId=11205&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
请实现一个函数用来匹配包括 '.' 和 '\*' 的正则表达式。模式中的字符 '.' 表示任意一个字符，而 '\*' 表示它前面的字符可以出现任意次（包含 0 次）。
|
||||
|
||||
在本题中，匹配是指字符串的所有字符匹配整个模式。例如，字符串 "aaa" 与模式 "a.a" 和 "ab\*ac\*a" 匹配，但是与 "aa.a" 和 "ab\*a" 均不匹配。
|
||||
|
||||
## 解题思路
|
||||
|
||||
应该注意到，'.' 是用来当做一个任意字符，而 '\*' 是用来重复前面的字符。这两个的作用不同，不能把 '.' 的作用和 '\*' 进行类比，从而把它当成重复前面字符一次。
|
||||
|
||||
```java
|
||||
public boolean match(char[] str, char[] pattern) {
|
||||
|
||||
int m = str.length, n = pattern.length;
|
||||
boolean[][] dp = new boolean[m + 1][n + 1];
|
||||
|
||||
dp[0][0] = true;
|
||||
for (int i = 1; i <= n; i++)
|
||||
if (pattern[i - 1] == '*')
|
||||
dp[0][i] = dp[0][i - 2];
|
||||
|
||||
for (int i = 1; i <= m; i++)
|
||||
for (int j = 1; j <= n; j++)
|
||||
if (str[i - 1] == pattern[j - 1] || pattern[j - 1] == '.')
|
||||
dp[i][j] = dp[i - 1][j - 1];
|
||||
else if (pattern[j - 1] == '*')
|
||||
if (pattern[j - 2] == str[i - 1] || pattern[j - 2] == '.') {
|
||||
dp[i][j] |= dp[i][j - 1]; // a* counts as single a
|
||||
dp[i][j] |= dp[i - 1][j]; // a* counts as multiple a
|
||||
dp[i][j] |= dp[i][j - 2]; // a* counts as empty
|
||||
} else
|
||||
dp[i][j] = dp[i][j - 2]; // a* only counts as empty
|
||||
|
||||
return dp[m][n];
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
56
docs/notes/20. 表示数值的字符串.md
Normal file
56
docs/notes/20. 表示数值的字符串.md
Normal file
@ -0,0 +1,56 @@
|
||||
# 20. 表示数值的字符串
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/6f8c901d091949a5837e24bb82a731f2?tpId=13&tqId=11206&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
```
|
||||
true
|
||||
|
||||
"+100"
|
||||
"5e2"
|
||||
"-123"
|
||||
"3.1416"
|
||||
"-1E-16"
|
||||
```
|
||||
|
||||
```
|
||||
false
|
||||
|
||||
"12e"
|
||||
"1a3.14"
|
||||
"1.2.3"
|
||||
"+-5"
|
||||
"12e+4.3"
|
||||
```
|
||||
|
||||
|
||||
## 解题思路
|
||||
|
||||
使用正则表达式进行匹配。
|
||||
|
||||
```html
|
||||
[] ： 字符集合
|
||||
() ： 分组
|
||||
? ： 重复 0 ~ 1 次
|
||||
+ ： 重复 1 ~ n 次
|
||||
* ： 重复 0 ~ n 次
|
||||
. ： 任意字符
|
||||
\\. ： 转义后的 .
|
||||
\\d ： 数字
|
||||
```
|
||||
|
||||
```java
|
||||
public boolean isNumeric(char[] str) {
|
||||
if (str == null || str.length == 0)
|
||||
return false;
|
||||
return new String(str).matches("[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?");
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
67
docs/notes/21. 调整数组顺序使奇数位于偶数前面.md
Normal file
67
docs/notes/21. 调整数组顺序使奇数位于偶数前面.md
Normal file
@ -0,0 +1,67 @@
|
||||
# 21. 调整数组顺序使奇数位于偶数前面
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/beb5aa231adc45b2a5dcc5b62c93f593?tpId=13&tqId=11166&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
需要保证奇数和奇数，偶数和偶数之间的相对位置不变，这和书本不太一样。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/d03a2efa-ef19-4c96-97e8-ff61df8061d3.png" width="200px"> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
方法一：创建一个新数组，时间复杂度 O(N)，空间复杂度 O(N)。
|
||||
|
||||
```java
|
||||
public void reOrderArray(int[] nums) {
|
||||
// 奇数个数
|
||||
int oddCnt = 0;
|
||||
for (int x : nums)
|
||||
if (!isEven(x))
|
||||
oddCnt++;
|
||||
int[] copy = nums.clone();
|
||||
int i = 0, j = oddCnt;
|
||||
for (int num : copy) {
|
||||
if (num % 2 == 1)
|
||||
nums[i++] = num;
|
||||
else
|
||||
nums[j++] = num;
|
||||
}
|
||||
}
|
||||
|
||||
private boolean isEven(int x) {
|
||||
return x % 2 == 0;
|
||||
}
|
||||
```
|
||||
|
||||
方法二：使用冒泡思想，每次都将当前偶数上浮到当前最右边。时间复杂度 O(N<sup>2</sup>)，空间复杂度 O(1)，时间换空间。
|
||||
|
||||
```java
|
||||
public void reOrderArray(int[] nums) {
|
||||
int N = nums.length;
|
||||
for (int i = N - 1; i > 0; i--) {
|
||||
for (int j = 0; j < i; j++) {
|
||||
if (isEven(nums[j]) && !isEven(nums[j + 1])) {
|
||||
swap(nums, j, j + 1);
|
||||
}
|
||||
}
|
||||
}
|
||||
}
|
||||
|
||||
private boolean isEven(int x) {
|
||||
return x % 2 == 0;
|
||||
}
|
||||
|
||||
private void swap(int[] nums, int i, int j) {
|
||||
int t = nums[i];
|
||||
nums[i] = nums[j];
|
||||
nums[j] = t;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
34
docs/notes/22. 链表中倒数第 K 个结点.md
Normal file
34
docs/notes/22. 链表中倒数第 K 个结点.md
Normal file
@ -0,0 +1,34 @@
|
||||
# 22. 链表中倒数第 K 个结点
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/529d3ae5a407492994ad2a246518148a?tpId=13&tqId=11167&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 解题思路
|
||||
|
||||
设链表的长度为 N。设置两个指针 P1 和 P2，先让 P1 移动 K 个节点，则还有 N - K 个节点可以移动。此时让 P1 和 P2 同时移动，可以知道当 P1 移动到链表结尾时，P2 移动到第 N - K 个节点处，该位置就是倒数第 K 个节点。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/6b504f1f-bf76-4aab-a146-a9c7a58c2029.png" width="500"/> </div><br>
|
||||
|
||||
```java
|
||||
public ListNode FindKthToTail(ListNode head, int k) {
|
||||
if (head == null)
|
||||
return null;
|
||||
ListNode P1 = head;
|
||||
while (P1 != null && k-- > 0)
|
||||
P1 = P1.next;
|
||||
if (k > 0)
|
||||
return null;
|
||||
ListNode P2 = head;
|
||||
while (P1 != null) {
|
||||
P1 = P1.next;
|
||||
P2 = P2.next;
|
||||
}
|
||||
return P2;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
50
docs/notes/23. 链表中环的入口结点.md
Normal file
50
docs/notes/23. 链表中环的入口结点.md
Normal file
@ -0,0 +1,50 @@
|
||||
# 23. 链表中环的入口结点
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/253d2c59ec3e4bc68da16833f79a38e4?tpId=13&tqId=11208&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
一个链表中包含环，请找出该链表的环的入口结点。要求不能使用额外的空间。
|
||||
|
||||
## 解题思路
|
||||
|
||||
使用双指针，一个快指针 fast 每次移动两个节点，一个慢指针 slow 每次移动一个节点。因为存在环，所以两个指针必定相遇在环中的某个节点上。
|
||||
|
||||
假设环入口节点为 y1，相遇所在节点为 z1。
|
||||
|
||||
假设快指针 fast 在圈内绕了 N 圈，则总路径长度为 x+Ny+(N-1)z。z 为 (N-1) 倍是因为快慢指针最后已经在 z1 节点相遇了，后面就不需要再走了。
|
||||
|
||||
而慢指针 slow 总路径长度为 x+y。
|
||||
|
||||
因为快指针是慢指针的两倍，因此 x+Ny+(N-1)z = 2(x+y)。
|
||||
|
||||
我们要找的是环入口节点 y1，也可以看成寻找长度 x 的值，因此我们先将上面的等值分解为和 x 有关：x=(N-2)y+(N-1)z。
|
||||
|
||||
上面的等值没有很强的规律，但是我们可以发现 y+z 就是圆环的总长度，因此我们将上面的等式再分解：x=(N-2)(y+z)+z。这个等式左边是从起点x1 到环入口节点 y1 的长度，而右边是在圆环中走过 (N-2) 圈，再从相遇点 z1 再走过长度为 z 的长度。此时我们可以发现如果让两个指针同时从起点 x1 和相遇点 z1 开始，每次只走过一个距离，那么最后他们会在环入口节点相遇。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/bb7fc182-98c2-4860-8ea3-630e27a5f29f.png" width="500"/> </div><br>
|
||||
|
||||
```java
|
||||
public ListNode EntryNodeOfLoop(ListNode pHead) {
|
||||
if (pHead == null || pHead.next == null)
|
||||
return null;
|
||||
ListNode slow = pHead, fast = pHead;
|
||||
do {
|
||||
fast = fast.next.next;
|
||||
slow = slow.next;
|
||||
} while (slow != fast);
|
||||
fast = pHead;
|
||||
while (slow != fast) {
|
||||
slow = slow.next;
|
||||
fast = fast.next;
|
||||
}
|
||||
return slow;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
43
docs/notes/24. 反转链表.md
Normal file
43
docs/notes/24. 反转链表.md
Normal file
@ -0,0 +1,43 @@
|
||||
# 24. 反转链表
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/75e878df47f24fdc9dc3e400ec6058ca?tpId=13&tqId=11168&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 解题思路
|
||||
|
||||
### 递归
|
||||
|
||||
```java
|
||||
public ListNode ReverseList(ListNode head) {
|
||||
if (head == null || head.next == null)
|
||||
return head;
|
||||
ListNode next = head.next;
|
||||
head.next = null;
|
||||
ListNode newHead = ReverseList(next);
|
||||
next.next = head;
|
||||
return newHead;
|
||||
}
|
||||
```
|
||||
|
||||
### 迭代
|
||||
|
||||
使用头插法。
|
||||
|
||||
```java
|
||||
public ListNode ReverseList(ListNode head) {
|
||||
ListNode newList = new ListNode(-1);
|
||||
while (head != null) {
|
||||
ListNode next = head.next;
|
||||
head.next = newList.next;
|
||||
newList.next = head;
|
||||
head = next;
|
||||
}
|
||||
return newList.next;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
58
docs/notes/25. 合并两个排序的链表.md
Normal file
58
docs/notes/25. 合并两个排序的链表.md
Normal file
@ -0,0 +1,58 @@
|
||||
# 25. 合并两个排序的链表
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/d8b6b4358f774294a89de2a6ac4d9337?tpId=13&tqId=11169&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/c094d2bc-ec75-444b-af77-d369dfb6b3b4.png" width="400"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
### 递归
|
||||
|
||||
```java
|
||||
public ListNode Merge(ListNode list1, ListNode list2) {
|
||||
if (list1 == null)
|
||||
return list2;
|
||||
if (list2 == null)
|
||||
return list1;
|
||||
if (list1.val <= list2.val) {
|
||||
list1.next = Merge(list1.next, list2);
|
||||
return list1;
|
||||
} else {
|
||||
list2.next = Merge(list1, list2.next);
|
||||
return list2;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
### 迭代
|
||||
|
||||
```java
|
||||
public ListNode Merge(ListNode list1, ListNode list2) {
|
||||
ListNode head = new ListNode(-1);
|
||||
ListNode cur = head;
|
||||
while (list1 != null && list2 != null) {
|
||||
if (list1.val <= list2.val) {
|
||||
cur.next = list1;
|
||||
list1 = list1.next;
|
||||
} else {
|
||||
cur.next = list2;
|
||||
list2 = list2.next;
|
||||
}
|
||||
cur = cur.next;
|
||||
}
|
||||
if (list1 != null)
|
||||
cur.next = list1;
|
||||
if (list2 != null)
|
||||
cur.next = list2;
|
||||
return head.next;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
34
docs/notes/26. 树的子结构.md
Normal file
34
docs/notes/26. 树的子结构.md
Normal file
@ -0,0 +1,34 @@
|
||||
# 26. 树的子结构
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/6e196c44c7004d15b1610b9afca8bd88?tpId=13&tqId=11170&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/84a5b15a-86c5-4d8e-9439-d9fd5a4699a1.jpg" width="450"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
public boolean HasSubtree(TreeNode root1, TreeNode root2) {
|
||||
if (root1 == null || root2 == null)
|
||||
return false;
|
||||
return isSubtreeWithRoot(root1, root2) || HasSubtree(root1.left, root2) || HasSubtree(root1.right, root2);
|
||||
}
|
||||
|
||||
private boolean isSubtreeWithRoot(TreeNode root1, TreeNode root2) {
|
||||
if (root2 == null)
|
||||
return true;
|
||||
if (root1 == null)
|
||||
return false;
|
||||
if (root1.val != root2.val)
|
||||
return false;
|
||||
return isSubtreeWithRoot(root1.left, root2.left) && isSubtreeWithRoot(root1.right, root2.right);
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
32
docs/notes/27. 二叉树的镜像.md
Normal file
32
docs/notes/27. 二叉树的镜像.md
Normal file
@ -0,0 +1,32 @@
|
||||
# 27. 二叉树的镜像
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/564f4c26aa584921bc75623e48ca3011?tpId=13&tqId=11171&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/0c12221f-729e-4c22-b0ba-0dfc909f8adf.jpg" width="300"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
public void Mirror(TreeNode root) {
|
||||
if (root == null)
|
||||
return;
|
||||
swap(root);
|
||||
Mirror(root.left);
|
||||
Mirror(root.right);
|
||||
}
|
||||
|
||||
private void swap(TreeNode root) {
|
||||
TreeNode t = root.left;
|
||||
root.left = root.right;
|
||||
root.right = t;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
34
docs/notes/28. 对称的二叉树.md
Normal file
34
docs/notes/28. 对称的二叉树.md
Normal file
@ -0,0 +1,34 @@
|
||||
# 28. 对称的二叉树
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/ff05d44dfdb04e1d83bdbdab320efbcb?tpId=13&tqId=11211&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/0c12221f-729e-4c22-b0ba-0dfc909f8adf.jpg" width="300"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
boolean isSymmetrical(TreeNode pRoot) {
|
||||
if (pRoot == null)
|
||||
return true;
|
||||
return isSymmetrical(pRoot.left, pRoot.right);
|
||||
}
|
||||
|
||||
boolean isSymmetrical(TreeNode t1, TreeNode t2) {
|
||||
if (t1 == null && t2 == null)
|
||||
return true;
|
||||
if (t1 == null || t2 == null)
|
||||
return false;
|
||||
if (t1.val != t2.val)
|
||||
return false;
|
||||
return isSymmetrical(t1.left, t2.right) && isSymmetrical(t1.right, t2.left);
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
39
docs/notes/29. 顺时针打印矩阵.md
Normal file
39
docs/notes/29. 顺时针打印矩阵.md
Normal file
@ -0,0 +1,39 @@
|
||||
# 29. 顺时针打印矩阵
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/9b4c81a02cd34f76be2659fa0d54342a?tpId=13&tqId=11172&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
下图的矩阵顺时针打印结果为：1, 2, 3, 4, 8, 12, 16, 15, 14, 13, 9, 5, 6, 7, 11, 10
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/48517227-324c-4664-bd26-a2d2cffe2bfe.png" width="200px"> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
public ArrayList<Integer> printMatrix(int[][] matrix) {
|
||||
ArrayList<Integer> ret = new ArrayList<>();
|
||||
int r1 = 0, r2 = matrix.length - 1, c1 = 0, c2 = matrix[0].length - 1;
|
||||
while (r1 <= r2 && c1 <= c2) {
|
||||
for (int i = c1; i <= c2; i++)
|
||||
ret.add(matrix[r1][i]);
|
||||
for (int i = r1 + 1; i <= r2; i++)
|
||||
ret.add(matrix[i][c2]);
|
||||
if (r1 != r2)
|
||||
for (int i = c2 - 1; i >= c1; i--)
|
||||
ret.add(matrix[r2][i]);
|
||||
if (c1 != c2)
|
||||
for (int i = r2 - 1; i > r1; i--)
|
||||
ret.add(matrix[i][c1]);
|
||||
r1++; r2--; c1++; c2--;
|
||||
}
|
||||
return ret;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
58
docs/notes/3. 数组中重复的数字.md
Normal file
58
docs/notes/3. 数组中重复的数字.md
Normal file
@ -0,0 +1,58 @@
|
||||
# 3. 数组中重复的数字
|
||||
|
||||
## 题目链接
|
||||
|
||||
[牛客网](https://www.nowcoder.com/practice/623a5ac0ea5b4e5f95552655361ae0a8?tpId=13&tqId=11203&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
在一个长度为 n 的数组里的所有数字都在 0 到 n-1 的范围内。数组中某些数字是重复的，但不知道有几个数字是重复的，也不知道每个数字重复几次。请找出数组中任意一个重复的数字。
|
||||
|
||||
```html
|
||||
Input:
|
||||
{2, 3, 1, 0, 2, 5}
|
||||
|
||||
Output:
|
||||
2
|
||||
```
|
||||
|
||||
## 解题思路
|
||||
|
||||
要求时间复杂度 O(N)，空间复杂度 O(1)。因此不能使用排序的方法，也不能使用额外的标记数组。
|
||||
|
||||
对于这种数组元素在 [0, n-1] 范围内的问题，可以将值为 i 的元素调整到第 i 个位置上进行求解。本题要求找出重复的数字，因此在调整过程中，如果第 i 位置上已经有一个值为 i 的元素，就可以知道 i 值重复。
|
||||
|
||||
以 (2, 3, 1, 0, 2, 5) 为例，遍历到位置 4 时，该位置上的数为 2，但是第 2 个位置上已经有一个 2 的值了，因此可以知道 2 重复：
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/643b6f18-f933-4ac5-aa7a-e304dbd7fe49.gif" width="350px"> </div><br>
|
||||
|
||||
|
||||
```java
|
||||
public boolean duplicate(int[] nums, int length, int[] duplication) {
|
||||
if (nums == null || length <= 0)
|
||||
return false;
|
||||
for (int i = 0; i < length; i++) {
|
||||
while (nums[i] != i) {
|
||||
if (nums[i] == nums[nums[i]]) {
|
||||
duplication[0] = nums[i];
|
||||
return true;
|
||||
}
|
||||
swap(nums, i, nums[i]);
|
||||
}
|
||||
}
|
||||
return false;
|
||||
}
|
||||
|
||||
private void swap(int[] nums, int i, int j) {
|
||||
int t = nums[i];
|
||||
nums[i] = nums[j];
|
||||
nums[j] = t;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
39
docs/notes/30. 包含 min 函数的栈.md
Normal file
39
docs/notes/30. 包含 min 函数的栈.md
Normal file
@ -0,0 +1,39 @@
|
||||
# 30. 包含 min 函数的栈
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/4c776177d2c04c2494f2555c9fcc1e49?tpId=13&tqId=11173&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
定义栈的数据结构，请在该类型中实现一个能够得到栈最小元素的 min 函数。
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
private Stack<Integer> dataStack = new Stack<>();
|
||||
private Stack<Integer> minStack = new Stack<>();
|
||||
|
||||
public void push(int node) {
|
||||
dataStack.push(node);
|
||||
minStack.push(minStack.isEmpty() ? node : Math.min(minStack.peek(), node));
|
||||
}
|
||||
|
||||
public void pop() {
|
||||
dataStack.pop();
|
||||
minStack.pop();
|
||||
}
|
||||
|
||||
public int top() {
|
||||
return dataStack.peek();
|
||||
}
|
||||
|
||||
public int min() {
|
||||
return minStack.peek();
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
36
docs/notes/31. 栈的压入、弹出序列.md
Normal file
36
docs/notes/31. 栈的压入、弹出序列.md
Normal file
@ -0,0 +1,36 @@
|
||||
# 31. 栈的压入、弹出序列
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/d77d11405cc7470d82554cb392585106?tpId=13&tqId=11174&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
输入两个整数序列，第一个序列表示栈的压入顺序，请判断第二个序列是否为该栈的弹出顺序。假设压入栈的所有数字均不相等。
|
||||
|
||||
例如序列 1,2,3,4,5 是某栈的压入顺序，序列 4,5,3,2,1 是该压栈序列对应的一个弹出序列，但 4,3,5,1,2 就不可能是该压栈序列的弹出序列。
|
||||
|
||||
## 解题思路
|
||||
|
||||
使用一个栈来模拟压入弹出操作。
|
||||
|
||||
```java
|
||||
public boolean IsPopOrder(int[] pushSequence, int[] popSequence) {
|
||||
int n = pushSequence.length;
|
||||
Stack<Integer> stack = new Stack<>();
|
||||
for (int pushIndex = 0, popIndex = 0; pushIndex < n; pushIndex++) {
|
||||
stack.push(pushSequence[pushIndex]);
|
||||
while (popIndex < n && !stack.isEmpty()
|
||||
&& stack.peek() == popSequence[popIndex]) {
|
||||
stack.pop();
|
||||
popIndex++;
|
||||
}
|
||||
}
|
||||
return stack.isEmpty();
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
44
docs/notes/32.1 从上往下打印二叉树.md
Normal file
44
docs/notes/32.1 从上往下打印二叉树.md
Normal file
@ -0,0 +1,44 @@
|
||||
# 32.1 从上往下打印二叉树
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/7fe2212963db4790b57431d9ed259701?tpId=13&tqId=11175&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
从上往下打印出二叉树的每个节点，同层节点从左至右打印。
|
||||
|
||||
例如，以下二叉树层次遍历的结果为：1,2,3,4,5,6,7
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/d5e838cf-d8a2-49af-90df-1b2a714ee676.jpg" width="250"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
使用队列来进行层次遍历。
|
||||
|
||||
不需要使用两个队列分别存储当前层的节点和下一层的节点，因为在开始遍历一层的节点时，当前队列中的节点数就是当前层的节点数，只要控制遍历这么多节点数，就能保证这次遍历的都是当前层的节点。
|
||||
|
||||
```java
|
||||
public ArrayList<Integer> PrintFromTopToBottom(TreeNode root) {
|
||||
Queue<TreeNode> queue = new LinkedList<>();
|
||||
ArrayList<Integer> ret = new ArrayList<>();
|
||||
queue.add(root);
|
||||
while (!queue.isEmpty()) {
|
||||
int cnt = queue.size();
|
||||
while (cnt-- > 0) {
|
||||
TreeNode t = queue.poll();
|
||||
if (t == null)
|
||||
continue;
|
||||
ret.add(t.val);
|
||||
queue.add(t.left);
|
||||
queue.add(t.right);
|
||||
}
|
||||
}
|
||||
return ret;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
39
docs/notes/32.2 把二叉树打印成多行.md
Normal file
39
docs/notes/32.2 把二叉树打印成多行.md
Normal file
@ -0,0 +1,39 @@
|
||||
# 32.2 把二叉树打印成多行
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/445c44d982d04483b04a54f298796288?tpId=13&tqId=11213&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
和上题几乎一样。
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
|
||||
ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
|
||||
Queue<TreeNode> queue = new LinkedList<>();
|
||||
queue.add(pRoot);
|
||||
while (!queue.isEmpty()) {
|
||||
ArrayList<Integer> list = new ArrayList<>();
|
||||
int cnt = queue.size();
|
||||
while (cnt-- > 0) {
|
||||
TreeNode node = queue.poll();
|
||||
if (node == null)
|
||||
continue;
|
||||
list.add(node.val);
|
||||
queue.add(node.left);
|
||||
queue.add(node.right);
|
||||
}
|
||||
if (list.size() != 0)
|
||||
ret.add(list);
|
||||
}
|
||||
return ret;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
43
docs/notes/32.3 按之字形顺序打印二叉树.md
Normal file
43
docs/notes/32.3 按之字形顺序打印二叉树.md
Normal file
@ -0,0 +1,43 @@
|
||||
# 32.3 按之字形顺序打印二叉树
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/91b69814117f4e8097390d107d2efbe0?tpId=13&tqId=11212&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
请实现一个函数按照之字形打印二叉树，即第一行按照从左到右的顺序打印，第二层按照从右至左的顺序打印，第三行按照从左到右的顺序打印，其他行以此类推。
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
public ArrayList<ArrayList<Integer>> Print(TreeNode pRoot) {
|
||||
ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
|
||||
Queue<TreeNode> queue = new LinkedList<>();
|
||||
queue.add(pRoot);
|
||||
boolean reverse = false;
|
||||
while (!queue.isEmpty()) {
|
||||
ArrayList<Integer> list = new ArrayList<>();
|
||||
int cnt = queue.size();
|
||||
while (cnt-- > 0) {
|
||||
TreeNode node = queue.poll();
|
||||
if (node == null)
|
||||
continue;
|
||||
list.add(node.val);
|
||||
queue.add(node.left);
|
||||
queue.add(node.right);
|
||||
}
|
||||
if (reverse)
|
||||
Collections.reverse(list);
|
||||
reverse = !reverse;
|
||||
if (list.size() != 0)
|
||||
ret.add(list);
|
||||
}
|
||||
return ret;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
41
docs/notes/33. 二叉搜索树的后序遍历序列.md
Normal file
41
docs/notes/33. 二叉搜索树的后序遍历序列.md
Normal file
@ -0,0 +1,41 @@
|
||||
# 33. 二叉搜索树的后序遍历序列
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/a861533d45854474ac791d90e447bafd?tpId=13&tqId=11176&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历的结果。假设输入的数组的任意两个数字都互不相同。
|
||||
|
||||
例如，下图是后序遍历序列 1,3,2 所对应的二叉搜索树。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/13454fa1-23a8-4578-9663-2b13a6af564a.jpg" width="150"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
public boolean VerifySquenceOfBST(int[] sequence) {
|
||||
if (sequence == null || sequence.length == 0)
|
||||
return false;
|
||||
return verify(sequence, 0, sequence.length - 1);
|
||||
}
|
||||
|
||||
private boolean verify(int[] sequence, int first, int last) {
|
||||
if (last - first <= 1)
|
||||
return true;
|
||||
int rootVal = sequence[last];
|
||||
int cutIndex = first;
|
||||
while (cutIndex < last && sequence[cutIndex] <= rootVal)
|
||||
cutIndex++;
|
||||
for (int i = cutIndex; i < last; i++)
|
||||
if (sequence[i] < rootVal)
|
||||
return false;
|
||||
return verify(sequence, first, cutIndex - 1) && verify(sequence, cutIndex, last - 1);
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
43
docs/notes/34. 二叉树中和为某一值的路径.md
Normal file
43
docs/notes/34. 二叉树中和为某一值的路径.md
Normal file
@ -0,0 +1,43 @@
|
||||
# 34. 二叉树中和为某一值的路径
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/b736e784e3e34731af99065031301bca?tpId=13&tqId=11177&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
输入一颗二叉树和一个整数，打印出二叉树中结点值的和为输入整数的所有路径。路径定义为从树的根结点开始往下一直到叶结点所经过的结点形成一条路径。
|
||||
|
||||
下图的二叉树有两条和为 22 的路径：10, 5, 7 和 10, 12
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/ed77b0e6-38d9-4a34-844f-724f3ffa2c12.jpg" width="200"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
private ArrayList<ArrayList<Integer>> ret = new ArrayList<>();
|
||||
|
||||
public ArrayList<ArrayList<Integer>> FindPath(TreeNode root, int target) {
|
||||
backtracking(root, target, new ArrayList<>());
|
||||
return ret;
|
||||
}
|
||||
|
||||
private void backtracking(TreeNode node, int target, ArrayList<Integer> path) {
|
||||
if (node == null)
|
||||
return;
|
||||
path.add(node.val);
|
||||
target -= node.val;
|
||||
if (target == 0 && node.left == null && node.right == null) {
|
||||
ret.add(new ArrayList<>(path));
|
||||
} else {
|
||||
backtracking(node.left, target, path);
|
||||
backtracking(node.right, target, path);
|
||||
}
|
||||
path.remove(path.size() - 1);
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
74
docs/notes/35. 复杂链表的复制.md
Normal file
74
docs/notes/35. 复杂链表的复制.md
Normal file
@ -0,0 +1,74 @@
|
||||
# 35. 复杂链表的复制
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/f836b2c43afc4b35ad6adc41ec941dba?tpId=13&tqId=11178&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
输入一个复杂链表（每个节点中有节点值，以及两个指针，一个指向下一个节点，另一个特殊指针指向任意一个节点），返回结果为复制后复杂链表的 head。
|
||||
|
||||
```java
|
||||
public class RandomListNode {
|
||||
int label;
|
||||
RandomListNode next = null;
|
||||
RandomListNode random = null;
|
||||
|
||||
RandomListNode(int label) {
|
||||
this.label = label;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/66a01953-5303-43b1-8646-0c77b825e980.png" width="300"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
第一步，在每个节点的后面插入复制的节点。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/dfd5d3f8-673c-486b-8ecf-d2082107b67b.png" width="600"/> </div><br>
|
||||
|
||||
第二步，对复制节点的 random 链接进行赋值。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/cafbfeb8-7dfe-4c0a-a3c9-750eeb824068.png" width="600"/> </div><br>
|
||||
|
||||
第三步，拆分。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/e151b5df-5390-4365-b66e-b130cd253c12.png" width="600"/> </div><br>
|
||||
|
||||
```java
|
||||
public RandomListNode Clone(RandomListNode pHead) {
|
||||
if (pHead == null)
|
||||
return null;
|
||||
// 插入新节点
|
||||
RandomListNode cur = pHead;
|
||||
while (cur != null) {
|
||||
RandomListNode clone = new RandomListNode(cur.label);
|
||||
clone.next = cur.next;
|
||||
cur.next = clone;
|
||||
cur = clone.next;
|
||||
}
|
||||
// 建立 random 链接
|
||||
cur = pHead;
|
||||
while (cur != null) {
|
||||
RandomListNode clone = cur.next;
|
||||
if (cur.random != null)
|
||||
clone.random = cur.random.next;
|
||||
cur = clone.next;
|
||||
}
|
||||
// 拆分
|
||||
cur = pHead;
|
||||
RandomListNode pCloneHead = pHead.next;
|
||||
while (cur.next != null) {
|
||||
RandomListNode next = cur.next;
|
||||
cur.next = next.next;
|
||||
cur = next;
|
||||
}
|
||||
return pCloneHead;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
41
docs/notes/36. 二叉搜索树与双向链表.md
Normal file
41
docs/notes/36. 二叉搜索树与双向链表.md
Normal file
@ -0,0 +1,41 @@
|
||||
# 36. 二叉搜索树与双向链表
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/947f6eb80d944a84850b0538bf0ec3a5?tpId=13&tqId=11179&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
输入一棵二叉搜索树，将该二叉搜索树转换成一个排序的双向链表。要求不能创建任何新的结点，只能调整树中结点指针的指向。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/05a08f2e-9914-4a77-92ef-aebeaecf4f66.jpg" width="400"/> </div><br>
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
private TreeNode pre = null;
|
||||
private TreeNode head = null;
|
||||
|
||||
public TreeNode Convert(TreeNode root) {
|
||||
inOrder(root);
|
||||
return head;
|
||||
}
|
||||
|
||||
private void inOrder(TreeNode node) {
|
||||
if (node == null)
|
||||
return;
|
||||
inOrder(node.left);
|
||||
node.left = pre;
|
||||
if (pre != null)
|
||||
pre.right = node;
|
||||
pre = node;
|
||||
if (head == null)
|
||||
head = node;
|
||||
inOrder(node.right);
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
46
docs/notes/37. 序列化二叉树.md
Normal file
46
docs/notes/37. 序列化二叉树.md
Normal file
@ -0,0 +1,46 @@
|
||||
# 37. 序列化二叉树
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/cf7e25aa97c04cc1a68c8f040e71fb84?tpId=13&tqId=11214&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
请实现两个函数，分别用来序列化和反序列化二叉树。
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
private String deserializeStr;
|
||||
|
||||
public String Serialize(TreeNode root) {
|
||||
if (root == null)
|
||||
return "#";
|
||||
return root.val + " " + Serialize(root.left) + " " + Serialize(root.right);
|
||||
}
|
||||
|
||||
public TreeNode Deserialize(String str) {
|
||||
deserializeStr = str;
|
||||
return Deserialize();
|
||||
}
|
||||
|
||||
private TreeNode Deserialize() {
|
||||
if (deserializeStr.length() == 0)
|
||||
return null;
|
||||
int index = deserializeStr.indexOf(" ");
|
||||
String node = index == -1 ? deserializeStr : deserializeStr.substring(0, index);
|
||||
deserializeStr = index == -1 ? "" : deserializeStr.substring(index + 1);
|
||||
if (node.equals("#"))
|
||||
return null;
|
||||
int val = Integer.valueOf(node);
|
||||
TreeNode t = new TreeNode(val);
|
||||
t.left = Deserialize();
|
||||
t.right = Deserialize();
|
||||
return t;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
47
docs/notes/38. 字符串的排列.md
Normal file
47
docs/notes/38. 字符串的排列.md
Normal file
@ -0,0 +1,47 @@
|
||||
# 38. 字符串的排列
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/fe6b651b66ae47d7acce78ffdd9a96c7?tpId=13&tqId=11180&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
输入一个字符串，按字典序打印出该字符串中字符的所有排列。例如输入字符串 abc，则打印出由字符 a, b, c 所能排列出来的所有字符串 abc, acb, bac, bca, cab 和 cba。
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
private ArrayList<String> ret = new ArrayList<>();
|
||||
|
||||
public ArrayList<String> Permutation(String str) {
|
||||
if (str.length() == 0)
|
||||
return ret;
|
||||
char[] chars = str.toCharArray();
|
||||
Arrays.sort(chars);
|
||||
backtracking(chars, new boolean[chars.length], new StringBuilder());
|
||||
return ret;
|
||||
}
|
||||
|
||||
private void backtracking(char[] chars, boolean[] hasUsed, StringBuilder s) {
|
||||
if (s.length() == chars.length) {
|
||||
ret.add(s.toString());
|
||||
return;
|
||||
}
|
||||
for (int i = 0; i < chars.length; i++) {
|
||||
if (hasUsed[i])
|
||||
continue;
|
||||
if (i != 0 && chars[i] == chars[i - 1] && !hasUsed[i - 1]) /* 保证不重复 */
|
||||
continue;
|
||||
hasUsed[i] = true;
|
||||
s.append(chars[i]);
|
||||
backtracking(chars, hasUsed, s);
|
||||
s.deleteCharAt(s.length() - 1);
|
||||
hasUsed[i] = false;
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
34
docs/notes/39. 数组中出现次数超过一半的数字.md
Normal file
34
docs/notes/39. 数组中出现次数超过一半的数字.md
Normal file
@ -0,0 +1,34 @@
|
||||
# 39. 数组中出现次数超过一半的数字
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/e8a1b01a2df14cb2b228b30ee6a92163?tpId=13&tqId=11181&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 解题思路
|
||||
|
||||
多数投票问题，可以利用 Boyer-Moore Majority Vote Algorithm 来解决这个问题，使得时间复杂度为 O(N)。
|
||||
|
||||
使用 cnt 来统计一个元素出现的次数，当遍历到的元素和统计元素相等时，令 cnt++，否则令 cnt--。如果前面查找了 i 个元素，且 cnt == 0，说明前 i 个元素没有 majority，或者有 majority，但是出现的次数少于 i / 2 ，因为如果多于 i / 2 的话 cnt 就一定不会为 0 。此时剩下的 n - i 个元素中，majority 的数目依然多于 (n - i) / 2，因此继续查找就能找出 majority。
|
||||
|
||||
```java
|
||||
public int MoreThanHalfNum_Solution(int[] nums) {
|
||||
int majority = nums[0];
|
||||
for (int i = 1, cnt = 1; i < nums.length; i++) {
|
||||
cnt = nums[i] == majority ? cnt + 1 : cnt - 1;
|
||||
if (cnt == 0) {
|
||||
majority = nums[i];
|
||||
cnt = 1;
|
||||
}
|
||||
}
|
||||
int cnt = 0;
|
||||
for (int val : nums)
|
||||
if (val == majority)
|
||||
cnt++;
|
||||
return cnt > nums.length / 2 ? majority : 0;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
56
docs/notes/4. 二维数组中的查找.md
Normal file
56
docs/notes/4. 二维数组中的查找.md
Normal file
@ -0,0 +1,56 @@
|
||||
# 4. 二维数组中的查找
|
||||
|
||||
## 题目链接
|
||||
|
||||
[牛客网](https://www.nowcoder.com/practice/abc3fe2ce8e146608e868a70efebf62e?tpId=13&tqId=11154&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
给定一个二维数组，其每一行从左到右递增排序，从上到下也是递增排序。给定一个数，判断这个数是否在该二维数组中。
|
||||
|
||||
```html
|
||||
Consider the following matrix:
|
||||
[
|
||||
[1, 4, 7, 11, 15],
|
||||
[2, 5, 8, 12, 19],
|
||||
[3, 6, 9, 16, 22],
|
||||
[10, 13, 14, 17, 24],
|
||||
[18, 21, 23, 26, 30]
|
||||
]
|
||||
|
||||
Given target = 5, return true.
|
||||
Given target = 20, return false.
|
||||
```
|
||||
|
||||
## 解题思路
|
||||
|
||||
要求时间复杂度 O(M + N)，空间复杂度 O(1)。其中 M 为行数，N 为 列数。
|
||||
|
||||
该二维数组中的一个数，小于它的数一定在其左边，大于它的数一定在其下边。因此，从右上角开始查找，就可以根据 target 和当前元素的大小关系来缩小查找区间，当前元素的查找区间为左下角的所有元素。
|
||||
|
||||
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/35a8c711-0dc0-4613-95f3-be96c6c6e104.gif" width="400px"> </div><br>
|
||||
|
||||
```java
|
||||
public boolean Find(int target, int[][] matrix) {
|
||||
if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
|
||||
return false;
|
||||
int rows = matrix.length, cols = matrix[0].length;
|
||||
int r = 0, c = cols - 1; // 从右上角开始
|
||||
while (r <= rows - 1 && c >= 0) {
|
||||
if (target == matrix[r][c])
|
||||
return true;
|
||||
else if (target > matrix[r][c])
|
||||
r++;
|
||||
else
|
||||
c--;
|
||||
}
|
||||
return false;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
88
docs/notes/40. 最小的 K 个数.md
Normal file
88
docs/notes/40. 最小的 K 个数.md
Normal file
@ -0,0 +1,88 @@
|
||||
# 40. 最小的 K 个数
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/6a296eb82cf844ca8539b57c23e6e9bf?tpId=13&tqId=11182&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 解题思路
|
||||
|
||||
### 快速选择
|
||||
|
||||
- 复杂度：O(N) + O(1)
|
||||
- 只有当允许修改数组元素时才可以使用
|
||||
|
||||
快速排序的 partition() 方法，会返回一个整数 j 使得 a[l..j-1] 小于等于 a[j]，且 a[j+1..h] 大于等于 a[j]，此时 a[j] 就是数组的第 j 大元素。可以利用这个特性找出数组的第 K 个元素，这种找第 K 个元素的算法称为快速选择算法。
|
||||
|
||||
```java
|
||||
public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k) {
|
||||
ArrayList<Integer> ret = new ArrayList<>();
|
||||
if (k > nums.length || k <= 0)
|
||||
return ret;
|
||||
findKthSmallest(nums, k - 1);
|
||||
/* findKthSmallest 会改变数组，使得前 k 个数都是最小的 k 个数 */
|
||||
for (int i = 0; i < k; i++)
|
||||
ret.add(nums[i]);
|
||||
return ret;
|
||||
}
|
||||
|
||||
public void findKthSmallest(int[] nums, int k) {
|
||||
int l = 0, h = nums.length - 1;
|
||||
while (l < h) {
|
||||
int j = partition(nums, l, h);
|
||||
if (j == k)
|
||||
break;
|
||||
if (j > k)
|
||||
h = j - 1;
|
||||
else
|
||||
l = j + 1;
|
||||
}
|
||||
}
|
||||
|
||||
private int partition(int[] nums, int l, int h) {
|
||||
int p = nums[l]; /* 切分元素 */
|
||||
int i = l, j = h + 1;
|
||||
while (true) {
|
||||
while (i != h && nums[++i] < p) ;
|
||||
while (j != l && nums[--j] > p) ;
|
||||
if (i >= j)
|
||||
break;
|
||||
swap(nums, i, j);
|
||||
}
|
||||
swap(nums, l, j);
|
||||
return j;
|
||||
}
|
||||
|
||||
private void swap(int[] nums, int i, int j) {
|
||||
int t = nums[i];
|
||||
nums[i] = nums[j];
|
||||
nums[j] = t;
|
||||
}
|
||||
```
|
||||
|
||||
### 大小为 K 的最小堆
|
||||
|
||||
- 复杂度：O(NlogK) + O(K)
|
||||
- 特别适合处理海量数据
|
||||
|
||||
应该使用大顶堆来维护最小堆，而不能直接创建一个小顶堆并设置一个大小，企图让小顶堆中的元素都是最小元素。
|
||||
|
||||
维护一个大小为 K 的最小堆过程如下：在添加一个元素之后，如果大顶堆的大小大于 K，那么需要将大顶堆的堆顶元素去除。
|
||||
|
||||
```java
|
||||
public ArrayList<Integer> GetLeastNumbers_Solution(int[] nums, int k) {
|
||||
if (k > nums.length || k <= 0)
|
||||
return new ArrayList<>();
|
||||
PriorityQueue<Integer> maxHeap = new PriorityQueue<>((o1, o2) -> o2 - o1);
|
||||
for (int num : nums) {
|
||||
maxHeap.add(num);
|
||||
if (maxHeap.size() > k)
|
||||
maxHeap.poll();
|
||||
}
|
||||
return new ArrayList<>(maxHeap);
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
47
docs/notes/41.1 数据流中的中位数.md
Normal file
47
docs/notes/41.1 数据流中的中位数.md
Normal file
@ -0,0 +1,47 @@
|
||||
# 41.1 数据流中的中位数
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/9be0172896bd43948f8a32fb954e1be1?tpId=13&tqId=11216&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
如何得到一个数据流中的中位数？如果从数据流中读出奇数个数值，那么中位数就是所有数值排序之后位于中间的数值。如果从数据流中读出偶数个数值，那么中位数就是所有数值排序之后中间两个数的平均值。
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
/* 大顶堆，存储左半边元素 */
|
||||
private PriorityQueue<Integer> left = new PriorityQueue<>((o1, o2) -> o2 - o1);
|
||||
/* 小顶堆，存储右半边元素，并且右半边元素都大于左半边 */
|
||||
private PriorityQueue<Integer> right = new PriorityQueue<>();
|
||||
/* 当前数据流读入的元素个数 */
|
||||
private int N = 0;
|
||||
|
||||
public void Insert(Integer val) {
|
||||
/* 插入要保证两个堆存于平衡状态 */
|
||||
if (N % 2 == 0) {
|
||||
/* N 为偶数的情况下插入到右半边。
|
||||
* 因为右半边元素都要大于左半边，但是新插入的元素不一定比左半边元素来的大，
|
||||
* 因此需要先将元素插入左半边，然后利用左半边为大顶堆的特点，取出堆顶元素即为最大元素，此时插入右半边 */
|
||||
left.add(val);
|
||||
right.add(left.poll());
|
||||
} else {
|
||||
right.add(val);
|
||||
left.add(right.poll());
|
||||
}
|
||||
N++;
|
||||
}
|
||||
|
||||
public Double GetMedian() {
|
||||
if (N % 2 == 0)
|
||||
return (left.peek() + right.peek()) / 2.0;
|
||||
else
|
||||
return (double) right.peek();
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
32
docs/notes/41.2 字符流中第一个不重复的字符.md
Normal file
32
docs/notes/41.2 字符流中第一个不重复的字符.md
Normal file
@ -0,0 +1,32 @@
|
||||
# 41.2 字符流中第一个不重复的字符
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/00de97733b8e4f97a3fb5c680ee10720?tpId=13&tqId=11207&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
请实现一个函数用来找出字符流中第一个只出现一次的字符。例如，当从字符流中只读出前两个字符 "go" 时，第一个只出现一次的字符是 "g"。当从该字符流中读出前六个字符“google" 时，第一个只出现一次的字符是 "l"。
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
private int[] cnts = new int[256];
|
||||
private Queue<Character> queue = new LinkedList<>();
|
||||
|
||||
public void Insert(char ch) {
|
||||
cnts[ch]++;
|
||||
queue.add(ch);
|
||||
while (!queue.isEmpty() && cnts[queue.peek()] > 1)
|
||||
queue.poll();
|
||||
}
|
||||
|
||||
public char FirstAppearingOnce() {
|
||||
return queue.isEmpty() ? '#' : queue.peek();
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
31
docs/notes/42. 连续子数组的最大和.md
Normal file
31
docs/notes/42. 连续子数组的最大和.md
Normal file
@ -0,0 +1,31 @@
|
||||
# 42. 连续子数组的最大和
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/459bd355da1549fa8a49e350bf3df484?tpId=13&tqId=11183&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 题目描述
|
||||
|
||||
{6, -3, -2, 7, -15, 1, 2, 2}，连续子数组的最大和为 8（从第 0 个开始，到第 3 个为止）。
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
public int FindGreatestSumOfSubArray(int[] nums) {
|
||||
if (nums == null || nums.length == 0)
|
||||
return 0;
|
||||
int greatestSum = Integer.MIN_VALUE;
|
||||
int sum = 0;
|
||||
for (int val : nums) {
|
||||
sum = sum <= 0 ? val : sum + val;
|
||||
greatestSum = Math.max(greatestSum, sum);
|
||||
}
|
||||
return greatestSum;
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
25
docs/notes/43. 从 1 到 n 整数中 1 出现的次数.md
Normal file
25
docs/notes/43. 从 1 到 n 整数中 1 出现的次数.md
Normal file
@ -0,0 +1,25 @@
|
||||
# 43. 从 1 到 n 整数中 1 出现的次数
|
||||
|
||||
[NowCoder](https://www.nowcoder.com/practice/bd7f978302044eee894445e244c7eee6?tpId=13&tqId=11184&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
||||
|
||||
## 解题思路
|
||||
|
||||
```java
|
||||
public int NumberOf1Between1AndN_Solution(int n) {
|
||||
int cnt = 0;
|
||||
for (int m = 1; m <= n; m *= 10) {
|
||||
int a = n / m, b = n % m;
|
||||
cnt += (a + 8) / 10 * m + (a % 10 == 1 ? b + 1 : 0);
|
||||
}
|
||||
return cnt;
|
||||
}
|
||||
```
|
||||
|
||||
> [Leetcode : 233. Number of Digit One](https://leetcode.com/problems/number-of-digit-one/discuss/64381/4+-lines-O(log-n)-C++JavaPython)
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
|
||||
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
|
||||
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