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notes/算法 - 其它.md
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# 汉诺塔
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/69d6c38d-1dec-4f72-ae60-60dbc10e9d15.png" width="300"/> </div><br>
<div align="center"> <img src="pics/69d6c38d-1dec-4f72-ae60-60dbc10e9d15.png" width="300"/> </div><br>
有三个柱子，分别为 from、buffer、to。需要将 from 上的圆盘全部移动到 to 上，并且要保证小圆盘始终在大圆盘上。
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① 将 n-1 个圆盘从 from -> buffer
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/f9240aa1-8d48-4959-b28a-7ca45c3e4d91.png" width="300"/> </div><br>
<div align="center"> <img src="pics/f9240aa1-8d48-4959-b28a-7ca45c3e4d91.png" width="300"/> </div><br>
② 将 1 个圆盘从 from -> to
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/f579cab0-3d49-4d00-8e14-e9e1669d0f9f.png" width="300"/> </div><br>
<div align="center"> <img src="pics/f579cab0-3d49-4d00-8e14-e9e1669d0f9f.png" width="300"/> </div><br>
③ 将 n-1 个圆盘从 buffer -> to
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/d02f74dd-8e33-4f3c-bf29-53203a06695a.png" width="300"/> </div><br>
<div align="center"> <img src="pics/d02f74dd-8e33-4f3c-bf29-53203a06695a.png" width="300"/> </div><br>
如果只有一个圆盘，那么只需要进行一次移动操作。
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生成编码时，从根节点出发，向左遍历则添加二进制位 0，向右则添加二进制位 1，直到遍历到叶子节点，叶子节点代表的字符的编码就是这个路径编码。
<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/8edc5164-810b-4cc5-bda8-2a2c98556377.jpg" width="300"/> </div><br>
<div align="center"> <img src="pics/8edc5164-810b-4cc5-bda8-2a2c98556377.jpg" width="300"/> </div><br>
```java
public class Huffman {