141 lines
4.8 KiB
Markdown
141 lines
4.8 KiB
Markdown
<!-- GFM-TOC -->
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* [汉诺塔](#汉诺塔)
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* [哈夫曼编码](#哈夫曼编码)
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<!-- GFM-TOC -->
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# 汉诺塔
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<div align="center"> <img src="pics/54f1e052-0596-4b5e-833c-e80d75bf3f9b.png" width="300"/> </div><br>
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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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这是一个经典的递归问题，分为三步求解：
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① 将 n-1 个圆盘从 from -> buffer
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<div align="center"> <img src="pics/8587132a-021d-4f1f-a8ec-5a9daa7157a7.png" width="300"/> </div><br>
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② 将 1 个圆盘从 from -> to
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<div align="center"> <img src="pics/2861e923-4862-4526-881c-15529279d49c.png" width="300"/> </div><br>
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③ 将 n-1 个圆盘从 buffer -> to
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<div align="center"> <img src="pics/1c4e8185-8153-46b6-bd5a-288b15feeae6.png" width="300"/> </div><br>
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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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从上面的讨论可以知道，a<sub>n</sub> = 2 * a<sub>n-1</sub> + 1，显然 a<sub>n</sub> = 2<sup>n</sup> - 1，n 个圆盘需要移动 2<sup>n</sup> - 1 次。
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```java
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public class Hanoi {
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public static void move(int n, String from, String buffer, String to) {
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if (n == 1) {
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System.out.println("from " + from + " to " + to);
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return;
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}
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move(n - 1, from, to, buffer);
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move(1, from, buffer, to);
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move(n - 1, buffer, from, to);
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}
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public static void main(String[] args) {
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Hanoi.move(3, "H1", "H2", "H3");
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}
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}
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```
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```html
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from H1 to H3
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from H1 to H2
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from H3 to H2
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from H1 to H3
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from H2 to H1
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from H2 to H3
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from H1 to H3
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```
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# 哈夫曼编码
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根据数据出现的频率对数据进行编码，从而压缩原始数据。
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例如对于一个文本文件，其中各种字符出现的次数如下：
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- a : 10
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- b : 20
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- c : 40
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- d : 80
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可以将每种字符转换成二进制编码，例如将 a 转换为 00，b 转换为 01，c 转换为 10，d 转换为 11。这是最简单的一种编码方式，没有考虑各个字符的权值（出现频率）。而哈夫曼编码采用了贪心策略，使出现频率最高的字符的编码最短，从而保证整体的编码长度最短。
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首先生成一颗哈夫曼树，每次生成过程中选取频率最少的两个节点，生成一个新节点作为它们的父节点，并且新节点的频率为两个节点的和。选取频率最少的原因是，生成过程使得先选取的节点位于树的更低层，那么需要的编码长度更长，频率更少可以使得总编码长度更少。
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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。
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<div align="center"> <img src="pics/3ff4f00a-2321-48fd-95f4-ce6001332151.png" width="400"/> </div><br>
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```java
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public class Huffman {
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private class Node implements Comparable<Node> {
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char ch;
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int freq;
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boolean isLeaf;
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Node left, right;
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public Node(char ch, int freq) {
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this.ch = ch;
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this.freq = freq;
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isLeaf = true;
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}
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public Node(Node left, Node right, int freq) {
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this.left = left;
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this.right = right;
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this.freq = freq;
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isLeaf = false;
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}
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@Override
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public int compareTo(Node o) {
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return this.freq - o.freq;
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}
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}
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public Map<Character, String> encode(Map<Character, Integer> frequencyForChar) {
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PriorityQueue<Node> priorityQueue = new PriorityQueue<>();
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for (Character c : frequencyForChar.keySet()) {
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priorityQueue.add(new Node(c, frequencyForChar.get(c)));
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}
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while (priorityQueue.size() != 1) {
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Node node1 = priorityQueue.poll();
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Node node2 = priorityQueue.poll();
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priorityQueue.add(new Node(node1, node2, node1.freq + node2.freq));
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}
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return encode(priorityQueue.poll());
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}
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private Map<Character, String> encode(Node root) {
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Map<Character, String> encodingForChar = new HashMap<>();
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encode(root, "", encodingForChar);
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return encodingForChar;
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}
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private void encode(Node node, String encoding, Map<Character, String> encodingForChar) {
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if (node.isLeaf) {
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encodingForChar.put(node.ch, encoding);
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return;
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}
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encode(node.left, encoding + '0', encodingForChar);
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encode(node.right, encoding + '1', encodingForChar);
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}
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}
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```
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</br><div align="center">公众号 CyC2018，专注于核心基础知识分享、求职指导、技术成长。在公众号后台回复 ziliao 可领取复习大纲，帮你理清复习重点。</div></br>
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<div align="center"><img width="180px" src="https://cyc-1256109796.cos.ap-guangzhou.myqcloud.com/%E5%85%AC%E4%BC%97%E5%8F%B7.jpg"></img></div>
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