1972 lines
56 KiB
Markdown
1972 lines
56 KiB
Markdown
<!-- GFM-TOC -->
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* [1. 字符串组合](#1-字符串组合)
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* [2. 整数组合求和](#2-整数组合求和)
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* [3. 数组中重复的数字](#3-数组中重复的数字)
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* [4. 二维数组中的查找](#4-二维数组中的查找)
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* [5. 替换空格](#5-替换空格)
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* [6. 从尾到头打印链表](#6-从尾到头打印链表)
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* [7. 重建二叉树](#7-重建二叉树)
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* [8. 二叉树的下一个结点](#8-二叉树的下一个结点)
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* [9. 用两个栈实现队列](#9-用两个栈实现队列)
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* [10.1 斐波那契数列](#101-斐波那契数列)
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* [10.2 跳台阶](#102-跳台阶)
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* [10.3 矩形覆盖](#103-矩形覆盖)
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* [10.4 变态跳台阶](#104-变态跳台阶)
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* [11. 旋转数组的最小数字](#11-旋转数组的最小数字)
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* [12. 矩阵中的路径](#12-矩阵中的路径)
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* [13. 机器人的运动范围](#13-机器人的运动范围)
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* [14. 剪绳子](#14-剪绳子)
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* [15. 二进制中 1 的个数](#15-二进制中-1-的个数)
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* [16. 数值的整数次方](#16-数值的整数次方)
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* [17. 打印从 1 到最大的 n 位数](#17-打印从-1-到最大的-n-位数)
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* [18.1 在 O(1) 时间内删除链表节点](#181-在-o1-时间内删除链表节点)
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* [18.2 删除链表中重复的结点](#182-删除链表中重复的结点)
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* [19. 正则表达式匹配](#19-正则表达式匹配)
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* [20. 表示数值的字符串](#20-表示数值的字符串)
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* [21. 调整数组顺序使奇数位于偶数前面](#21-调整数组顺序使奇数位于偶数前面)
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* [22. 链表中倒数第 K 个结点](#22-链表中倒数第-k-个结点)
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* [23. 链表中环的入口结点](#23-链表中环的入口结点)
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* [24. 反转链表](#24-反转链表)
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* [25. 合并两个排序的链表](#25-合并两个排序的链表)
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* [参考文献](#参考文献)
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<!-- GFM-TOC -->
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# 1. 字符串组合
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## 题目描述
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给定三种类型的小球P、Q、R，每种小球的数量分别为np、nq、nr个。现在想讲这些小球排成一条直线，但是不允许相同类型的小球相邻，问有多少种排序方法。
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如若np=2，nq=1,nr=1则共有6种排列方式：PQRP、QPRP、PRQP、PRPQ、RPQP以及PQPR。如果无法组合出合适的结果，则输出0.
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### 输入
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```code
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一行以空格相隔的三个数，分别表示为np、nq、nr。
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```
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### 输出
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```code
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排列方法的数量。
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```
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### 样例输入
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```code
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2 1 1
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```
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### 样例输出
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```code
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6
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```
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## 解题思路
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本题采用一种比较直接的方式进行解题，分为如下步骤：
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1. 求解给定P、Q、R个数的时候的全排列，提供`Python`提供的`itertools.permutations`来实现，此时肯定有很多重复。
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2. 去掉重复的情况，通过`Python`提供的`set`来实现。
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3. 通过遍历找出相邻元素重复的串。
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4. 求两个集合的差集，即为答案。
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```python
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import itertools
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np, nq, nr = [int(k) for k in raw_input().split(" ")]
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count = 0
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result = []
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for i in itertools.permutations("P"*np + "Q"*nq + "R"*nr,np + nq + nr):
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result.append(''.join(i))
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result_same = []
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for element in list(set(result)):
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for j in range(1, len(element)):
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if element[j-1] == element[j]:
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result_same.append(element)
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ret_list = list(set(result)^set(result_same))
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print len(ret_list)
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```
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# 2. 整数组合求和
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## 题目描述
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小米之家是成人糖果店。里面有很多便宜，好用，好玩的产品。中秋节到了，小米之家想给米粉准备一些固定金额大礼包。对于给定的一个金额，需要判断能不能用
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不同种产品（一种产品在礼包最多出现一次）组合出来这个金额。聪明的你来帮帮米家的小伙伴吧。
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### 输入
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```code
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输入N（N是正整数， N < = 200）
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输入N个价格p（正整数， p <= 10000）用单空格分割
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输入金额M（M是正整数，M <= 100000）
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```
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### 输出
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```code
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能组合出来输出1
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否则输出0
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```
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### 样例输入
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```code
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6
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99 199 1999 10000 39 1499
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10238
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```
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### 样例输出
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```code
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1
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```
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## 解题思路
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本题采用一种比较直接的方式进行解题，分为如下步骤：
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1. 对输入np进行排序，方便后面快速结束。
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2. 求解给定np时候的0~n个组合的和，提供`Python`提供的`itertools.permutations`来实现。
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3. 如果较小的数相加已经大于目标，可以提前跳出本次循环。
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```python
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import itertools
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n = int(raw_input())
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np = [int(k) for k in raw_input().split(" ")]
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np.sort()
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sum_np = int(raw_input())
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flag = 0
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for i in range(n):
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for j in itertools.permutations(np, i):
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if sum(j) > sum_np:
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continue;
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if sum(j) == sum_np:
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flag = 1;
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break;
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print flag
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```
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# 3. 数组中重复的数字
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[NowCoder](https://www.nowcoder.com/practice/623a5ac0ea5b4e5f95552655361ae0a8?tpId=13&tqId=11203&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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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。例如，如果输入长度为 7 的数组 {2, 3, 1, 0, 2, 5}，那么对应的输出是第一个重复的数字 2。
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要求复杂度为 O(N) + O(1)，也就是时间复杂度 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。牛客网讨论区这一题的首票答案使用 nums[i] + length 来将元素标记，这么做会有加法溢出问题。
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## 解题思路
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这种数组元素在 [0, n-1] 范围内的问题，可以将值为 i 的元素放到第 i 个位置上。
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以 (2, 3, 1, 0, 2, 5) 为例：
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```text-html-basic
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position-0 : (2,3,1,0,2,5) // 2 <-> 1
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(1,3,2,0,2,5) // 1 <-> 3
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(3,1,2,0,2,5) // 3 <-> 0
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(0,1,2,3,2,5) // already in position
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position-1 : (0,1,2,3,2,5) // already in position
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position-2 : (0,1,2,3,2,5) // already in position
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position-3 : (0,1,2,3,2,5) // already in position
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position-4 : (0,1,2,3,2,5) // nums[i] == nums[nums[i]], exit
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```
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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。
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```java
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public boolean duplicate(int[] nums, int length, int[] duplication) {
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if (nums == null || length <= 0)
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return false;
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for (int i = 0; i < length; i++) {
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while (nums[i] != i) {
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if (nums[i] == nums[nums[i]]) {
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duplication[0] = nums[i];
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return true;
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}
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swap(nums, i, nums[i]);
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}
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}
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return false;
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}
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private void swap(int[] nums, int i, int j) {
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int t = nums[i]; nums[i] = nums[j]; nums[j] = t;
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}
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```
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```python
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nums = [int(k) for k in raw_input().split(" ")]
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print nums
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def duplicate(nums):
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if len(nums) <= 0:
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return -1, False
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for i in range(len(nums)):
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while nums[i] != i:
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if nums[i] == nums[nums[i]]:
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return nums[i], True
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index = nums[i]
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nums[i], nums[index] = nums[index], nums[i]
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# nums[i], nums[nums[i]] = nums[nums[i]], nums[i]
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return -1, False
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print duplicate(nums)
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```
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# 4. 二维数组中的查找
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[NowCoder](https://www.nowcoder.com/practice/abc3fe2ce8e146608e868a70efebf62e?tpId=13&tqId=11154&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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在一个二维数组中，每一行都按照从左到右递增的顺序排序，每一列都按照从上到下递增的顺序排序。请完成一个函数，输入这样的一个二维数组和一个整数，判断数组中是否含有该整数。
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```html
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Consider the following matrix:
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[
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[1, 4, 7, 11, 15],
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[2, 5, 8, 12, 19],
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[3, 6, 9, 16, 22],
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[10, 13, 14, 17, 24],
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[18, 21, 23, 26, 30]
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]
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Given target = 5, return true.
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Given target = 20, return false.
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```
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## 解题思路
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从右上角开始查找。矩阵中的一个数，它左边的数都比它小，下边的数都比它大。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。
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复杂度：O(M + N) + O(1)
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当前元素的查找区间为左下角的所有元素，例如元素 12 的查找区间如下：
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<div align="center"> <img src="../pics//f94389e9-55b1-4f49-9d37-00ed05900ae0.png" width="250"/> </div><br>
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```java
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public boolean Find(int target, int[][] matrix) {
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if (matrix == null || matrix.length == 0 || matrix[0].length == 0)
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return false;
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int rows = matrix.length, cols = matrix[0].length;
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int r = 0, c = cols - 1; // 从右上角开始
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while (r <= rows - 1 && c >= 0) {
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if (target == matrix[r][c])
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return true;
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else if (target > matrix[r][c])
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r++;
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else
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c--;
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}
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return false;
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}
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```
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```python
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target = int(input())
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# nums = [[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]]
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nums = eval(input())
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def find(target, matrix):
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if len(matrix) == 0 or len(matrix[0]) == 0:
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return False
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rows, cols = len(matrix), len(matrix[0])
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r, c = 0, cols - 1
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while r <= rows - 1 and c >= 0:
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if target == matrix[r][c]:
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return True
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elif target > matrix[r][c]:
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r += 1
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else:
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c -= 1
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return False
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print (find(target, nums))
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```
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# 5. 替换空格
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[NowCoder](https://www.nowcoder.com/practice/4060ac7e3e404ad1a894ef3e17650423?tpId=13&tqId=11155&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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将一个字符串中的空格替换成 "%20"。
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```text
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Input:
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"We Are Happy"
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Output:
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"We%20Are%20Happy"
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```
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## 解题思路
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在字符串尾部填充任意字符，使得字符串的长度等于替换之后的长度。因为一个空格要替换成三个字符（%20），因此当遍历到一个空格时，需要在尾部填充两个任意字符。
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令 P1 指向字符串原来的末尾位置，P2 指向字符串现在的末尾位置。P1 和 P2从后向前遍历，当 P1 遍历到一个空格时，就需要令 P2 指向的位置依次填充 02%（注意是逆序的），否则就填充上 P1 指向字符的值。
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从后向前遍是为了在改变 P2 所指向的内容时，不会影响到 P1 遍历原来字符串的内容。
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```java
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public String replaceSpace(StringBuffer str) {
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int P1 = str.length() - 1;
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for (int i = 0; i < P1 + 1; i++)
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if (str.charAt(i) == ' ')
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str.append(" ");
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int P2 = str.length() - 1;
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while (P1 >= 0 && P2 > P1) {
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char c = str.charAt(P1--);
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if (c == ' ') {
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str.setCharAt(P2--, '0');
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str.setCharAt(P2--, '2');
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str.setCharAt(P2--, '%');
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} else {
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str.setCharAt(P2--, c);
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}
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}
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return str.toString();
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}
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```
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```python
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target = input()
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print (target.replace(" ", "%20"))
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```
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# 6. 从尾到头打印链表
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[NowCoder](https://www.nowcoder.com/practice/d0267f7f55b3412ba93bd35cfa8e8035?tpId=13&tqId=11156&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
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## 题目描述
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输入链表的第一个节点，从尾到头反过来打印出每个结点的值。
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<div align="center"> <img src="../pics//d99dc9e2-197c-4085-813d-7195da1c6762.png" width="300"/> </div><br>
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## 解题思路
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### 使用栈
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```java
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public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
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Stack<Integer> stack = new Stack<>();
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while (listNode != null) {
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stack.add(listNode.val);
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listNode = listNode.next;
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}
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ArrayList<Integer> ret = new ArrayList<>();
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while (!stack.isEmpty())
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ret.add(stack.pop());
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return ret;
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}
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```
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### 使用递归
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```java
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public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
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ArrayList<Integer> ret = new ArrayList<>();
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if (listNode != null) {
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ret.addAll(printListFromTailToHead(listNode.next));
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ret.add(listNode.val);
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}
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return ret;
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}
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```
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### 使用头插法
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利用链表头插法为逆序的特点。
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头结点和第一个节点的区别：
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- 头结点是在头插法中使用的一个额外节点，这个节点不存储值；
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- 第一个节点就是链表的第一个真正存储值的节点。
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```java
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public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
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// 头插法构建逆序链表
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ListNode head = new ListNode(-1);
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while (listNode != null) {
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ListNode memo = listNode.next;
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listNode.next = head.next;
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head.next = listNode;
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listNode = memo;
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}
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// 构建 ArrayList
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ArrayList<Integer> ret = new ArrayList<>();
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head = head.next;
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while (head != null) {
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ret.add(head.val);
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head = head.next;
|
||
}
|
||
return ret;
|
||
}
|
||
```
|
||
### 使用 Collections.reverse()
|
||
|
||
```java
|
||
public ArrayList<Integer> printListFromTailToHead(ListNode listNode) {
|
||
ArrayList<Integer> ret = new ArrayList<>();
|
||
while (listNode != null) {
|
||
ret.add(listNode.val);
|
||
listNode = listNode.next;
|
||
}
|
||
Collections.reverse(ret);
|
||
return ret;
|
||
}
|
||
```
|
||
|
||
```python
|
||
def printListFromTailToHead(listNode):
|
||
# write code here
|
||
l = list()
|
||
while listNode:
|
||
l.append(listNode.val)
|
||
listNode = listNode.next
|
||
l.reverse()
|
||
return l
|
||
```
|
||
# 7. 重建二叉树
|
||
|
||
[NowCoder](https://www.nowcoder.com/practice/8a19cbe657394eeaac2f6ea9b0f6fcf6?tpId=13&tqId=11157&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
|
||
|
||
## 题目描述
|
||
|
||
根据二叉树的前序遍历和中序遍历的结果，重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。
|
||
|
||
```html
|
||
preorder = [3,9,20,15,7]
|
||
inorder = [9,3,15,20,7]
|
||
```
|
||
|
||
<div align="center"> <img src="../pics//8a4c6ad4-a816-47d1-b93f-7ca4f78ab67a.png" width="250"/> </div><br>
|
||
|
||
## 解题思路
|
||
|
||
前序遍历的第一个值为根节点的值，使用这个值将中序遍历结果分成两部分，左部分为树的左子树中序遍历结果，右部分为树的右子树中序遍历的结果。
|
||
|
||
```java
|
||
// 缓存中序遍历数组每个值对应的索引
|
||
private Map<Integer, Integer> indexForInOrders = new HashMap<>();
|
||
|
||
public TreeNode reConstructBinaryTree(int[] pre, int[] in) {
|
||
for (int i = 0; i < in.length; i++)
|
||
indexForInOrders.put(in[i], i);
|
||
return reConstructBinaryTree(pre, 0, pre.length - 1, 0);
|
||
}
|
||
|
||
private TreeNode reConstructBinaryTree(int[] pre, int preL, int preR, int inL) {
|
||
if (preL > preR)
|
||
return null;
|
||
TreeNode root = new TreeNode(pre[preL]);
|
||
int inIndex = indexForInOrders.get(root.val);
|
||
int leftTreeSize = inIndex - inL;
|
||
root.left = reConstructBinaryTree(pre, preL + 1, preL + leftTreeSize, inL);
|
||
root.right = reConstructBinaryTree(pre, preL + leftTreeSize + 1, preR, inL + leftTreeSize + 1);
|
||
return root;
|
||
}
|
||
```
|
||
|
||
```python
|
||
# 返回构造的TreeNode根节点
|
||
def reConstructBinaryTree(self, pre, tin):
|
||
# write code here
|
||
if not pre or not tin:
|
||
return None
|
||
root = TreeNode(pre.pop(0))
|
||
index = tin.index(root.val)
|
||
root.left = self.reConstructBinaryTree(pre, tin[:index])
|
||
root.right = self.reConstructBinaryTree(pre, tin[index + 1:])
|
||
return root
|
||
```
|
||
|
||
# 8. 二叉树的下一个结点
|
||
|
||
[NowCoder](https://www.nowcoder.com/practice/9023a0c988684a53960365b889ceaf5e?tpId=13&tqId=11210&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
|
||
|
||
## 题目描述
|
||
|
||
给定一个二叉树和其中的一个结点，请找出中序遍历顺序的下一个结点并且返回。注意，树中的结点不仅包含左右子结点，同时包含指向父结点的指针。
|
||
|
||
```java
|
||
public class TreeLinkNode {
|
||
int val;
|
||
TreeLinkNode left = null;
|
||
TreeLinkNode right = null;
|
||
TreeLinkNode next = null;
|
||
|
||
TreeLinkNode(int val) {
|
||
this.val = val;
|
||
}
|
||
}
|
||
```
|
||
|
||
## 解题思路
|
||
|
||
① 如果一个节点的右子树不为空，那么该节点的下一个节点是右子树的最左节点；
|
||
|
||
<div align="center"> <img src="../pics//cb0ed469-27ab-471b-a830-648b279103c8.png" width="250"/> </div><br>
|
||
|
||
② 否则，向上找第一个左链接指向的树包含该节点的祖先节点。
|
||
|
||
<div align="center"> <img src="../pics//e143f6da-d114-4ba4-8712-f65299047fa2.png" width="250"/> </div><br>
|
||
|
||
```java
|
||
public TreeLinkNode GetNext(TreeLinkNode pNode) {
|
||
if (pNode.right != null) {
|
||
TreeLinkNode node = pNode.right;
|
||
while (node.left != null)
|
||
node = node.left;
|
||
return node;
|
||
} else {
|
||
while (pNode.next != null) {
|
||
TreeLinkNode parent = pNode.next;
|
||
if (parent.left == pNode)
|
||
return parent;
|
||
pNode = pNode.next;
|
||
}
|
||
}
|
||
return null;
|
||
}
|
||
```
|
||
|
||
```python
|
||
def GetNext(self, pNode):
|
||
# write code here
|
||
# pNode is None
|
||
if not pNode:
|
||
return pNode
|
||
if pNode.right:
|
||
node = pNode.right
|
||
while node.left:
|
||
node = node.left
|
||
return node
|
||
else:
|
||
while pNode.next:
|
||
parent = pNode.next
|
||
if parent.left == pNode:
|
||
return parent
|
||
pNode = pNode.next
|
||
# pNode not have the next node
|
||
return None
|
||
```
|
||
# 9. 用两个栈实现队列
|
||
|
||
[NowCoder](https://www.nowcoder.com/practice/54275ddae22f475981afa2244dd448c6?tpId=13&tqId=11158&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
|
||
|
||
## 题目描述
|
||
|
||
用两个栈来实现一个队列，完成队列的 Push 和 Pop 操作。
|
||
|
||
## 解题思路
|
||
|
||
in 栈用来处理入栈（push）操作，out 栈用来处理出栈（pop）操作。一个元素进入 in 栈之后，出栈的顺序被反转。当元素要出栈时，需要先进入 out 栈，此时元素出栈顺序再一次被反转，因此出栈顺序就和最开始入栈顺序是相同的，先进入的元素先退出，这就是队列的顺序。
|
||
|
||
<div align="center"> <img src="../pics//5acf7550-86c5-4c5b-b912-8ce70ef9c34e.png" width="400"/> </div><br>
|
||
|
||
```java
|
||
Stack<Integer> in = new Stack<Integer>();
|
||
Stack<Integer> out = new Stack<Integer>();
|
||
|
||
public void push(int node) {
|
||
in.push(node);
|
||
}
|
||
|
||
public int pop() throws Exception {
|
||
if (out.isEmpty())
|
||
while (!in.isEmpty())
|
||
out.push(in.pop());
|
||
|
||
if (out.isEmpty())
|
||
throw new Exception("queue is empty");
|
||
|
||
return out.pop();
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def __init__(self):
|
||
self.stack1 = []
|
||
self.stack2 = []
|
||
def push(self, node):
|
||
# write code here
|
||
self.stack1.append(node)
|
||
def pop(self):
|
||
# return xx
|
||
if self.stack2 == []:
|
||
while self.stack1:
|
||
self.stack2.append(self.stack1.pop())
|
||
return self.stack2.pop()
|
||
return self.stack2.pop()
|
||
```
|
||
# 10.1 斐波那契数列
|
||
|
||
[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)
|
||
|
||
## 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."/></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(10) 需要计算 f(9) 和 f(8)，计算 f(9) 需要计算 f(8) 和 f(7)，可以看到 f(8) 被重复计算了。
|
||
|
||
<div align="center"> <img src="../pics//faecea49-9974-40db-9821-c8636137df61.jpg" width="300"/> </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 项的值了。
|
||
|
||
```java
|
||
public class Solution {
|
||
private int[] fib = new int[40];
|
||
|
||
public Solution() {
|
||
fib[1] = 1;
|
||
fib[2] = 2;
|
||
for (int i = 2; i < fib.length; i++)
|
||
fib[i] = fib[i - 1] + fib[i - 2];
|
||
}
|
||
|
||
public int Fibonacci(int n) {
|
||
return fib[n];
|
||
}
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
|
||
def __init__(self):
|
||
self.fib = [0,1]
|
||
for i in range(2,40):
|
||
self.fib.append(self.fib[i-1]+self.fib[i-2])
|
||
def Fibonacci(self, n):
|
||
# write code here
|
||
return self.fib[n]
|
||
```
|
||
# 10.2 跳台阶
|
||
|
||
[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)
|
||
|
||
## 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。
|
||
|
||
## 解题思路
|
||
|
||
```java
|
||
public int JumpFloor(int n) {
|
||
if (n <= 2)
|
||
return n;
|
||
int pre2 = 1, pre1 = 2;
|
||
int result = 1;
|
||
for (int i = 2; i < n; i++) {
|
||
result = pre2 + pre1;
|
||
pre2 = pre1;
|
||
pre1 = result;
|
||
}
|
||
return result;
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def jumpFloor(self, number):
|
||
# write code here
|
||
if number == 0:
|
||
return number
|
||
pre1, pre2 = 1, 1
|
||
#result = 0
|
||
for i in range(number):
|
||
pre1, pre2 = pre2, pre1+pre2
|
||
return pre1
|
||
```
|
||
# 10.3 矩形覆盖
|
||
|
||
[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)
|
||
|
||
## 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？
|
||
|
||
## 解题思路
|
||
|
||
```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;
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def rectCover(self, number):
|
||
# write code here
|
||
if number == 0:
|
||
return number
|
||
pre1, pre2 = 1, 1
|
||
#result = 0
|
||
for i in range(number):
|
||
pre1, pre2 = pre2, pre1+pre2
|
||
return pre1
|
||
```
|
||
# 10.4 变态跳台阶
|
||
|
||
[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)
|
||
|
||
## 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。
|
||
|
||
## 解题思路
|
||
|
||
```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];
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def jumpFloorII(self, number):
|
||
# write code here
|
||
if number <= 0:
|
||
return 0
|
||
return pow(2, number-1)
|
||
```
|
||
# 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)
|
||
|
||
## 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。
|
||
|
||
例如数组 {3, 4, 5, 1, 2} 为 {1, 2, 3, 4, 5} 的一个旋转，该数组的最小值为 1。
|
||
|
||
## 解题思路
|
||
|
||
在一个有序数组中查找一个元素可以用二分查找，二分查找也称为折半查找，每次都能将查找区间减半，这种折半特性的算法时间复杂度都为 O(logN)。
|
||
|
||
本题可以修改二分查找算法进行求解：
|
||
|
||
- 当 nums[m] <= nums[h] 的情况下，说明解在 [l, m] 之间，此时令 h = m；
|
||
- 否则解在 [m + 1, h] 之间，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的话，那么就会出现一个特殊的情况：nums[l] == nums[m] == nums[h]，那么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];
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def minNumberInRotateArray(self, rotateArray):
|
||
# write code here
|
||
if len(rotateArray) == 0:
|
||
return 0
|
||
return min(rotateArray)
|
||
```
|
||
|
||
# 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)
|
||
|
||
## 题目描述
|
||
|
||
请设计一个函数，用来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左，向右，向上，向下移动一个格子。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="../pics//e31abb94-9201-4e06-9902-61101b92f475.png" width="300"/> </div><br>
|
||
|
||
## 解题思路
|
||
|
||
```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 i = 0, idx = 0; i < rows; i++)
|
||
for (int j = 0; j < cols; j++)
|
||
matrix[i][j] = array[idx++];
|
||
return matrix;
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def hasPath(self, matrix, rows, cols, path):
|
||
# write code here
|
||
for i in range(rows):
|
||
for j in range(cols):
|
||
if matrix[i*cols+j] == path[0]:
|
||
if self.find(list(matrix),rows,cols,path[1:],i,j):
|
||
return True
|
||
return False
|
||
def find(self,matrix,rows,cols,path,i,j):
|
||
if not path:
|
||
return True
|
||
matrix[i*cols+j]='0'
|
||
if j+1<cols and matrix[i*cols+j+1]==path[0]:
|
||
return self.find(matrix,rows,cols,path[1:],i,j+1)
|
||
elif j-1>=0 and matrix[i*cols+j-1]==path[0]:
|
||
return self.find(matrix,rows,cols,path[1:],i,j-1)
|
||
elif i+1<rows and matrix[(i+1)*cols+j]==path[0]:
|
||
return self.find(matrix,rows,cols,path[1:],i+1,j)
|
||
elif i-1>=0 and matrix[(i-1)*cols+j]==path[0]:
|
||
return self.find(matrix,rows,cols,path[1:],i-1,j)
|
||
else:
|
||
return False
|
||
|
||
|
||
|
||
def buildMatrix(self, matrix):
|
||
array = [[0]*self.cols for i in range(self.rows)]
|
||
for i in range(self.rows):
|
||
idx = 0
|
||
for j in range(self.cols):
|
||
array[i][j] = matrix[idx]
|
||
idx += 1
|
||
return array
|
||
```
|
||
# 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)
|
||
|
||
## 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 (35,37)，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
|
||
|
||
```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];
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def __init__(self):
|
||
self.acc = 0
|
||
|
||
def movingCount(self, threshold, rows, cols):
|
||
# write code here
|
||
self.threshold = threshold
|
||
self.rows = rows
|
||
self.cols = cols
|
||
self.board = [[0 for _ in range(cols)] for _ in range(rows)]
|
||
|
||
self.traverse(0,0)
|
||
return self.acc
|
||
|
||
def block(self, r, c):
|
||
s = sum(map(int,str(r)+str(c)))
|
||
return s>self.threshold
|
||
|
||
def traverse(self, r, c):
|
||
if not (0<=r<self.rows and 0<=c<self.cols): return
|
||
if self.board[r][c]!=0: return
|
||
if self.board[r][c]==-1 or self.block(r,c):
|
||
self.board[r][c]=-1
|
||
return
|
||
self.board[r][c] = 1
|
||
self.acc += 1
|
||
self.traverse(r+1,c)
|
||
self.traverse(r-1,c)
|
||
self.traverse(r,c+1)
|
||
self.traverse(r,c-1)
|
||
```
|
||
|
||
# 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) - 2(n - 2) = n - 5 >= 0。因此把长度大于 5 的绳子切成两段，令其中一段长度为 3 可以使得两段的乘积最大。
|
||
|
||
```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));
|
||
}
|
||
```
|
||
|
||
|
||
```python
|
||
def integerBreak(n):
|
||
if n < 2:
|
||
return 0
|
||
if n == 2:
|
||
return 1
|
||
if n == 3:
|
||
return 2
|
||
timesOf3 = n // 3
|
||
if n - timesOf3 * 3 == 1:
|
||
timesOf3 -= 1
|
||
timesOf2 = (n - timesOf3 * 3) // 2
|
||
return pow(3, timesOf3) * pow(2, timesOf2)
|
||
```
|
||
# 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)
|
||
|
||
## 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);
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def NumberOf1(self, n):
|
||
# write code here
|
||
if n<0:
|
||
n=n&0xFFFFFFFF #把负号去掉，如果负号在后面会陷入死循环
|
||
return bin(n).count('1')
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def NumberOf1(self, n):
|
||
# write code here
|
||
return sum([(n>>i & 1) for i in range(0,32)])
|
||
```
|
||
|
||
# 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)
|
||
|
||
## 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."/></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;
|
||
}
|
||
```
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def Power(self, base, exponent):
|
||
# write code here
|
||
flag = 0
|
||
if base == 0:
|
||
return False
|
||
if exponent == 0:
|
||
return 1
|
||
if exponent < 0:
|
||
flag = 1
|
||
result = 1
|
||
for i in range(abs(exponent)):
|
||
result *= base
|
||
if flag == 1:
|
||
result = 1 / result
|
||
return result
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def Power(self, base, exponent):
|
||
# write code here
|
||
return pow(base, exponent)
|
||
```
|
||
|
||
### 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];
|
||
}
|
||
```
|
||
# 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();
|
||
}
|
||
```
|
||
|
||
```python
|
||
def numbersByRecursion(n,largest,result):
|
||
def recursion(num,largest,result):
|
||
if num <= largest:
|
||
result.append(num)
|
||
return recursion(num+1,largest,result)
|
||
else:
|
||
return result
|
||
return recursion(n,largest,result)
|
||
|
||
|
||
|
||
|
||
n = 2
|
||
result = []
|
||
largest = pow(10,n)-1
|
||
re = numbersByRecursion(1, largest, result)
|
||
print (re)
|
||
```
|
||
# 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="../pics//27ff9548-edb6-4465-92c8-7e6386e0b185.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="../pics//280f7728-594f-4811-a03a-fa8d32c013da.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 || head.next == 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 {
|
||
ListNode cur = head;
|
||
while (cur.next != tobeDelete)
|
||
cur = cur.next;
|
||
cur.next = null;
|
||
}
|
||
return head;
|
||
}
|
||
```
|
||
|
||
|
||
### 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];
|
||
}
|
||
```
|
||
|
||
```python
|
||
class ListNode:
|
||
def __init__(self, x = None):
|
||
self.val = x
|
||
self.next = None
|
||
def __del__(self):
|
||
self.val = None
|
||
self.next = None
|
||
|
||
class Solution:
|
||
def list_generate(self, lst):
|
||
"""
|
||
生成链表
|
||
"""
|
||
if not lst:
|
||
return None
|
||
list_node = ListNode()
|
||
list_node.value = lst[0]
|
||
if len(lst) == 1:
|
||
list_node.next_node = None
|
||
else:
|
||
list_node.next_node = self.list_generate(lst[1:])
|
||
return list_node
|
||
|
||
def find_node(self, node, target_value):
|
||
"""
|
||
根据给定的目标值，找出指定节点的位置
|
||
非题目要求，只是为了测试验证
|
||
"""
|
||
if not target_value:
|
||
return False
|
||
while node:
|
||
if node.value == target_value:
|
||
return node
|
||
node = node.next_node
|
||
return False
|
||
|
||
def delete_node(self, head_node, del_node):
|
||
"""
|
||
删除指定节点
|
||
"""
|
||
if not (head_node and del_node):
|
||
return False
|
||
|
||
if del_node.next_node:
|
||
# 删除的节点不是尾节点，而且不是唯一一个节点
|
||
del_next_node = del_node.next_node
|
||
del_node.value = del_next_node.value
|
||
del_node.next_node = del_next_node.next_node
|
||
del_next_node.__del__()
|
||
del_next_node.__del__()
|
||
|
||
elif del_node == head_node:
|
||
# 唯一一个节点，删除头节点
|
||
head_node = None
|
||
del_node = None
|
||
|
||
else:
|
||
# 删除节点为尾节点
|
||
node = head_node
|
||
while node.next_node != del_node:
|
||
node = node.next_node
|
||
|
||
node.next_node = None
|
||
del_node = None
|
||
|
||
return head_node
|
||
|
||
|
||
|
||
if __name__ == '__main__':
|
||
solution = Solution()
|
||
head_node = solution.list_generate(['a', 'b', 'c', 'e', 'd', 'd'])
|
||
|
||
target_value = 'a'
|
||
target_node = solution.find_node(head_node, target_value)
|
||
if target_node:
|
||
print (target_node.value)
|
||
head_node = solution.delete_node(head_node, target_node)
|
||
|
||
|
||
node = head_node
|
||
if node:
|
||
while node:
|
||
print (node.value)
|
||
node = node.next_node
|
||
if node:
|
||
print ('->')
|
||
else:
|
||
print ('wrong')
|
||
```
|
||
# 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)
|
||
|
||
## 题目描述
|
||
|
||
<div align="center"> <img src="../pics//8433fbb2-c35c-45ef-831d-e3ca42aebd51.png" width="500"/> </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;
|
||
}
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
# class ListNode:
|
||
# def __init__(self, x):
|
||
# self.val = x
|
||
# self.next = None
|
||
class Solution:
|
||
def deleteDuplication(self, pHead):
|
||
# write code here
|
||
if not pHead or not pHead.next:
|
||
return pHead
|
||
next = pHead.next
|
||
if pHead.val == next.val:
|
||
while next and pHead.val == next.val:
|
||
next = next.next
|
||
return self.deleteDuplication(next)
|
||
else:
|
||
pHead.next = self.deleteDuplication(pHead.next)
|
||
return pHead
|
||
```
|
||
|
||
# 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)
|
||
|
||
## 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];
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
# s, pattern都是字符串
|
||
def match(self, s, pattern):
|
||
# write code here
|
||
if (len(s) == 0 and len(pattern) == 0):
|
||
return True
|
||
if (len(s) > 0 and len(pattern) == 0):
|
||
return False
|
||
if (len(pattern) > 1 and pattern[1] == '*'):
|
||
if (len(s) > 0 and (s[0] == pattern[0] or pattern[0] == '.')):
|
||
return (self.match(s, pattern[2:]) or self.match(s[1:], pattern[2:]) or self.match(s[1:], pattern))
|
||
else:
|
||
return self.match(s, pattern[2:])
|
||
if (len(s) > 0 and (pattern[0] == '.' or pattern[0] == s[0])):
|
||
return self.match(s[1:], pattern[1:])
|
||
return False
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
import re
|
||
class Solution:
|
||
# s, pattern都是字符串
|
||
def match(self, s, pattern):
|
||
# write code here
|
||
if s == None or pattern == None:
|
||
return False
|
||
if s == "" and pattern == "":
|
||
return True
|
||
if pattern == "" and s != "":
|
||
return False
|
||
|
||
res = re.compile(str(pattern)+"$")
|
||
result = res.match(str(s))
|
||
|
||
if result is None:
|
||
return False
|
||
else:
|
||
return True
|
||
```
|
||
# 20. 表示数值的字符串
|
||
|
||
[NowCoder](https://www.nowcoder.com/practice/6f8c901d091949a5837e24bb82a731f2?tpId=13&tqId=11206&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking)
|
||
|
||
## 题目描述
|
||
|
||
```html
|
||
true
|
||
|
||
"+100"
|
||
"5e2"
|
||
"-123"
|
||
"3.1416"
|
||
"-1E-16"
|
||
|
||
false
|
||
|
||
"12e"
|
||
"1a3.14"
|
||
"1.2.3"
|
||
"+-5"
|
||
"12e+4.3"
|
||
```
|
||
|
||
|
||
## 解题思路
|
||
|
||
使用正则表达式进行匹配。
|
||
|
||
```html
|
||
[] ： 字符集合
|
||
() ： 分组
|
||
? ： 重复 0 ~ 1
|
||
+ ： 重复 1 ~ n
|
||
* ： 重复 0 ~ n
|
||
. ： 任意字符
|
||
\\. ： 转义后的 .
|
||
\\d ： 数字
|
||
```
|
||
|
||
```java
|
||
public boolean isNumeric(char[] str) {
|
||
if (str == null || str.length == 0)
|
||
return false;
|
||
return new String(str).matches("[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?");
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
import re
|
||
class Solution:
|
||
# s字符串
|
||
def isNumeric(self, s):
|
||
# write code here
|
||
return re.match('[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?$', s)
|
||
```
|
||
|
||
# 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)
|
||
|
||
## 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。
|
||
|
||
## 解题思路
|
||
|
||
```java
|
||
public void reOrderArray(int[] nums) {
|
||
// 奇数个数
|
||
int oddCnt = 0;
|
||
for (int val : nums)
|
||
if (val % 2 == 1)
|
||
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;
|
||
}
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
import copy
|
||
|
||
class Solution:
|
||
def reOrderArray(self, array):
|
||
# write code here
|
||
oddCnt = 0
|
||
for val in array:
|
||
if val % 2 == 1:
|
||
oddCnt += 1
|
||
copyArray = copy.deepcopy(array)
|
||
i, j = 0, oddCnt
|
||
for num in array:
|
||
if num % 2 ==1:
|
||
copyArray[i] = num
|
||
i += 1
|
||
else:
|
||
copyArray[j] = num
|
||
j += 1
|
||
return copyArray
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
class Solution:
|
||
def reOrderArray(self, array):
|
||
# write code here
|
||
odd = [x for x in array if x%2]
|
||
even = [x for x in array if not (x%2)]
|
||
result = odd + even
|
||
return result
|
||
```
|
||
# 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)
|
||
|
||
## 解题思路
|
||
|
||
设链表的长度为 N。设两个指针 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 N - K 个节点处，该位置就是倒数第 K 个节点。
|
||
|
||
<div align="center"> <img src="../pics//ea2304ce-268b-4238-9486-4d8f8aea8ca4.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;
|
||
}
|
||
```
|
||
|
||
```python
|
||
class Solution:
|
||
def FindKthToTail(self, head, k):
|
||
# write code here
|
||
if head == None:
|
||
return None
|
||
p1 = head
|
||
while p1 and k > 0:
|
||
k -= 1
|
||
p1 = p1.next
|
||
if k > 0:
|
||
return None
|
||
p2 = head
|
||
while p1:
|
||
p1 = p1.next
|
||
p2 = p2.next
|
||
return p2
|
||
```
|
||
|
||
# 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)
|
||
|
||
## 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，一个指针 fast 每次移动两个节点，一个指针 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。假设相遇点在下图的 y6 位置，此时 fast 移动的节点数为 x+2y+z，slow 为 x+y，由于 fast 速度比 slow 快一倍，因此 x+2y+z=2(x+y)，得到 x=z。
|
||
|
||
在相遇点，slow 要到环的入口点还需要移动 z 个节点，如果让 fast 重新从头开始移动，并且速度变为每次移动一个节点，那么它到环入口点还需要移动 x 个节点。在上面已经推导出 x=z，因此 fast 和 slow 将在环入口点相遇。
|
||
|
||
<div align="center"> <img src="../pics//70fa1f83-dae7-456d-b94b-ce28963b2ba1.png"/> </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;
|
||
}
|
||
```
|
||
|
||
```python
|
||
class Solution:
|
||
def EntryNodeOfLoop(self, pHead):
|
||
# write code here
|
||
if not pHead:
|
||
return None
|
||
plist = []
|
||
|
||
while True:
|
||
plist.append(pHead)
|
||
pHead = pHead.next
|
||
if not pHead:
|
||
return None
|
||
if pHead in plist:
|
||
return pHead
|
||
```
|
||
# 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)
|
||
|
||
## 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;
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
# class ListNode:
|
||
# def __init__(self, x):
|
||
# self.val = x
|
||
# self.next = None
|
||
class Solution:
|
||
# 返回ListNode
|
||
def ReverseList(self, pHead):
|
||
# write code here
|
||
if pHead == None or pHead.next == None:
|
||
return pHead
|
||
next = pHead.next
|
||
pHead.next = None
|
||
newHead = self.ReverseList(next)
|
||
next.next = pHead
|
||
return newHead
|
||
```
|
||
|
||
### 迭代
|
||
|
||
```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;
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
# class ListNode:
|
||
# def __init__(self, x):
|
||
# self.val = x
|
||
# self.next = None
|
||
class Solution:
|
||
# 返回ListNode
|
||
def ReverseList(self, pHead):
|
||
# write code here
|
||
newList = ListNode(-1)
|
||
while pHead:
|
||
next = pHead.next
|
||
pHead.next = newList.next
|
||
newList.next = pHead
|
||
pHead = next
|
||
return newList.next
|
||
```
|
||
|
||
# 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)
|
||
|
||
## 题目描述
|
||
|
||
<div align="center"> <img src="../pics//43f2cafa-3568-4a89-a895-4725666b94a6.png" width="500"/> </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;
|
||
}
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
# class ListNode:
|
||
# def __init__(self, x):
|
||
# self.val = x
|
||
# self.next = None
|
||
class Solution:
|
||
# 返回合并后列表
|
||
def Merge(self, pHead1, pHead2):
|
||
# write code here
|
||
if pHead1 == None:
|
||
return pHead2
|
||
if pHead2 == None:
|
||
return pHead1
|
||
if pHead1.val <= pHead2.val:
|
||
pHead1.next = self.Merge(pHead1.next, pHead2)
|
||
return pHead1
|
||
else:
|
||
pHead2.next = self.Merge(pHead1, pHead2.next)
|
||
return pHead2
|
||
```
|
||
|
||
### 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;
|
||
}
|
||
```
|
||
|
||
```python
|
||
# -*- coding:utf-8 -*-
|
||
# class ListNode:
|
||
# def __init__(self, x):
|
||
# self.val = x
|
||
# self.next = None
|
||
class Solution:
|
||
# 返回合并后列表
|
||
def Merge(self, pHead1, pHead2):
|
||
# write code here
|
||
head = ListNode(-1)
|
||
cur = head
|
||
while pHead1 and pHead2:
|
||
if pHead1.val <= pHead2.val:
|
||
cur.next = pHead1
|
||
pHead1 = pHead1.next
|
||
else:
|
||
cur.next = pHead2
|
||
pHead2 = pHead2.next
|
||
cur = cur.next
|
||
if pHead1:
|
||
cur.next = pHead1
|
||
if pHead2:
|
||
cur.next = pHead2
|
||
return head.next
|
||
```
|
||
|
||
# 参考文献
|
||
|
||
- 何海涛. 剑指 Offer[M]. 电子工业出版社, 2012.
|