342 lines
11 KiB
Markdown
342 lines
11 KiB
Markdown
<p align="center">
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<a href="https://mp.weixin.qq.com/s/RsdcQ9umo09R6cfnwXZlrQ"><img src="https://img.shields.io/badge/PDF下载-代码随想录-blueviolet" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw"><img src="https://img.shields.io/badge/刷题-微信群-green" alt=""></a>
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<a href="https://space.bilibili.com/525438321"><img src="https://img.shields.io/badge/B站-代码随想录-orange" alt=""></a>
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<a href="https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ"><img src="https://img.shields.io/badge/知识星球-代码随想录-blue" alt=""></a>
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</p>
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<p align="center"><strong>欢迎大家<a href="https://mp.weixin.qq.com/s/tqCxrMEU-ajQumL1i8im9A">参与本项目</a>，贡献其他语言版本的代码，拥抱开源，让更多学习算法的小伙伴们收益！</strong></p>
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## 72. 编辑距离
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https://leetcode-cn.com/problems/edit-distance/
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给你两个单词 word1 和 word2，请你计算出将 word1 转换成 word2 所使用的最少操作数 。
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你可以对一个单词进行如下三种操作：
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* 插入一个字符
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* 删除一个字符
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* 替换一个字符
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示例 1：
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输入：word1 = "horse", word2 = "ros"
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输出：3
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解释：
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horse -> rorse (将 'h' 替换为 'r')
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rorse -> rose (删除 'r')
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rose -> ros (删除 'e')
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示例 2：
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输入：word1 = "intention", word2 = "execution"
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输出：5
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解释：
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intention -> inention (删除 't')
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inention -> enention (将 'i' 替换为 'e')
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enention -> exention (将 'n' 替换为 'x')
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exention -> exection (将 'n' 替换为 'c')
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exection -> execution (插入 'u')
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提示：
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* 0 <= word1.length, word2.length <= 500
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* word1 和 word2 由小写英文字母组成
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## 思路
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编辑距离终于来了，这道题目如果大家没有了解动态规划的话，会感觉超级复杂。
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编辑距离是用动规来解决的经典题目，这道题目看上去好像很复杂，但用动规可以很巧妙的算出最少编辑距离。
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接下来我依然使用动规五部曲，对本题做一个详细的分析：
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-----------------------
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### 1. 确定dp数组（dp table）以及下标的含义
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**dp[i][j] 表示以下标i-1为结尾的字符串word1，和以下标j-1为结尾的字符串word2，最近编辑距离为dp[i][j]**。
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这里在强调一下：为啥要表示下标i-1为结尾的字符串呢，为啥不表示下标i为结尾的字符串呢？
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用i来表示也可以！ 但我统一以下标i-1为结尾的字符串，在下面的递归公式中会容易理解一点。
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-----------------------
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### 2. 确定递推公式
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在确定递推公式的时候，首先要考虑清楚编辑的几种操作，整理如下：
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```
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if (word1[i - 1] == word2[j - 1])
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不操作
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if (word1[i - 1] != word2[j - 1])
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增
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删
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换
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```
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也就是如上4种情况。
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`if (word1[i - 1] == word2[j - 1])` 那么说明不用任何编辑，`dp[i][j]` 就应该是 `dp[i - 1][j - 1]`，即`dp[i][j] = dp[i - 1][j - 1];`
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此时可能有同学有点不明白，为啥要即`dp[i][j] = dp[i - 1][j - 1]`呢？
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那么就在回顾上面讲过的`dp[i][j]`的定义，`word1[i - 1]` 与 `word2[j - 1]`相等了，那么就不用编辑了，以下标i-2为结尾的字符串word1和以下标j-2为结尾的字符串`word2`的最近编辑距离`dp[i - 1][j - 1]`就是 `dp[i][j]`了。
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在下面的讲解中，如果哪里看不懂，就回想一下`dp[i][j]`的定义，就明白了。
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**在整个动规的过程中，最为关键就是正确理解`dp[i][j]`的定义！**
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`if (word1[i - 1] != word2[j - 1])`，此时就需要编辑了，如何编辑呢？
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* 操作一：word1增加一个元素，使其word1[i - 1]与word2[j - 1]相同，那么就是以下标i-2为结尾的word1 与 j-1为结尾的word2的最近编辑距离 加上一个增加元素的操作。
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即 `dp[i][j] = dp[i - 1][j] + 1;`
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* 操作二：word2添加一个元素，使其word1[i - 1]与word2[j - 1]相同，那么就是以下标i-1为结尾的word1 与 j-2为结尾的word2的最近编辑距离 加上一个增加元素的操作。
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即 `dp[i][j] = dp[i][j - 1] + 1;`
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这里有同学发现了，怎么都是添加元素，删除元素去哪了。
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**word2添加一个元素，相当于word1删除一个元素**，例如 `word1 = "ad" ，word2 = "a"`，`word1`删除元素`'d'`，`word2`添加一个元素`'d'`，变成`word1="a", word2="ad"`， 最终的操作数是一样！ dp数组如下图所示意的：
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```
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a a d
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+-----+-----+ +-----+-----+-----+
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| 0 | 1 | | 0 | 1 | 2 |
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+-----+-----+ ===> +-----+-----+-----+
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a | 1 | 0 | a | 1 | 0 | 1 |
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+-----+-----+ +-----+-----+-----+
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d | 2 | 1 |
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+-----+-----+
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```
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操作三：替换元素，`word1`替换`word1[i - 1]`，使其与`word2[j - 1]`相同，此时不用增加元素，那么以下标`i-2`为结尾的`word1` 与 `j-2`为结尾的`word2`的最近编辑距离 加上一个替换元素的操作。
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即 `dp[i][j] = dp[i - 1][j - 1] + 1;`
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综上，当 `if (word1[i - 1] != word2[j - 1])` 时取最小的，即：`dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;`
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递归公式代码如下：
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```CPP
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if (word1[i - 1] == word2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1];
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}
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else {
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dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
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}
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```
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---
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### 3. dp数组如何初始化
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再回顾一下dp[i][j]的定义：
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**dp[i][j] 表示以下标i-1为结尾的字符串word1，和以下标j-1为结尾的字符串word2，最近编辑距离为dp[i][j]**。
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那么dp[i][0] 和 dp[0][j] 表示什么呢？
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dp[i][0] ：以下标i-1为结尾的字符串word1，和空字符串word2，最近编辑距离为dp[i][0]。
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那么dp[i][0]就应该是i，对word1里的元素全部做删除操作，即：dp[i][0] = i;
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同理dp[0][j] = j;
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所以C++代码如下：
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```CPP
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for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
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for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
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```
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-----------------------
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### 4. 确定遍历顺序
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从如下四个递推公式：
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* `dp[i][j] = dp[i - 1][j - 1]`
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* `dp[i][j] = dp[i - 1][j - 1] + 1`
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* `dp[i][j] = dp[i][j - 1] + 1`
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* `dp[i][j] = dp[i - 1][j] + 1`
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可以看出dp[i][j]是依赖左方，上方和左上方元素的，如图：
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所以在dp矩阵中一定是从左到右从上到下去遍历。
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代码如下：
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```CPP
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for (int i = 1; i <= word1.size(); i++) {
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for (int j = 1; j <= word2.size(); j++) {
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if (word1[i - 1] == word2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1];
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}
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else {
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dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
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}
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}
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}
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```
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-----------------------
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### 5. 举例推导dp数组
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以示例1为例，输入：`word1 = "horse", word2 = "ros"`为例，dp矩阵状态图如下：
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以上动规五部分析完毕，C++代码如下：
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```CPP
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class Solution {
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public:
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int minDistance(string word1, string word2) {
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vector<vector<int>> dp(word1.size() + 1, vector<int>(word2.size() + 1, 0));
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for (int i = 0; i <= word1.size(); i++) dp[i][0] = i;
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for (int j = 0; j <= word2.size(); j++) dp[0][j] = j;
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for (int i = 1; i <= word1.size(); i++) {
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for (int j = 1; j <= word2.size(); j++) {
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if (word1[i - 1] == word2[j - 1]) {
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dp[i][j] = dp[i - 1][j - 1];
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}
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else {
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dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
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}
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}
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}
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return dp[word1.size()][word2.size()];
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}
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};
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```
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-----------------------
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## 其他语言版本
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Java：
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```java
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public int minDistance(String word1, String word2) {
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int m = word1.length();
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int n = word2.length();
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int[][] dp = new int[m + 1][n + 1];
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// 初始化
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for (int i = 1; i <= m; i++) {
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dp[i][0] = i;
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}
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for (int j = 1; j <= n; j++) {
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dp[0][j] = j;
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}
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for (int i = 1; i <= m; i++) {
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for (int j = 1; j <= n; j++) {
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// 因为dp数组有效位从1开始
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// 所以当前遍历到的字符串的位置为i-1 | j-1
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if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
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dp[i][j] = dp[i - 1][j - 1];
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} else {
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dp[i][j] = Math.min(Math.min(dp[i - 1][j - 1], dp[i][j - 1]), dp[i - 1][j]) + 1;
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}
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}
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}
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return dp[m][n];
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}
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```
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Python：
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```python
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class Solution:
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def minDistance(self, word1: str, word2: str) -> int:
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dp = [[0] * (len(word2)+1) for _ in range(len(word1)+1)]
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for i in range(len(word1)+1):
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dp[i][0] = i
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for j in range(len(word2)+1):
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dp[0][j] = j
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for i in range(1, len(word1)+1):
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for j in range(1, len(word2)+1):
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if word1[i-1] == word2[j-1]:
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dp[i][j] = dp[i-1][j-1]
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else:
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dp[i][j] = min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]) + 1
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return dp[-1][-1]
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```
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Go：
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```Go
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func minDistance(word1 string, word2 string) int {
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m, n := len(word1), len(word2)
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dp := make([][]int, m+1)
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for i := range dp {
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dp[i] = make([]int, n+1)
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}
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for i := 0; i < m+1; i++ {
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dp[i][0] = i // word1[i] 变成 word2[0], 删掉 word1[i], 需要 i 部操作
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}
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for j := 0; j < n+1; j++ {
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dp[0][j] = j // word1[0] 变成 word2[j], 插入 word1[j]，需要 j 部操作
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}
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for i := 1; i < m+1; i++ {
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for j := 1; j < n+1; j++ {
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if word1[i-1] == word2[j-1] {
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dp[i][j] = dp[i-1][j-1]
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} else { // Min(插入，删除，替换)
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dp[i][j] = Min(dp[i][j-1], dp[i-1][j], dp[i-1][j-1]) + 1
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}
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}
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}
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return dp[m][n]
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}
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func Min(args ...int) int {
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min := args[0]
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for _, item := range args {
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if item < min {
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min = item
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}
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}
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return min
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}
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```
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Javascript：
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```javascript
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const minDistance = (word1, word2) => {
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let dp = Array.from(Array(word1.length + 1), () => Array(word2.length+1).fill(0));
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for(let i = 1; i <= word1.length; i++) {
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dp[i][0] = i;
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}
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for(let j = 1; j <= word2.length; j++) {
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dp[0][j] = j;
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}
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for(let i = 1; i <= word1.length; i++) {
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for(let j = 1; j <= word2.length; j++) {
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if(word1[i-1] === word2[j-1]) {
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dp[i][j] = dp[i-1][j-1];
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} else {
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dp[i][j] = Math.min(dp[i-1][j] + 1, dp[i][j-1] + 1, dp[i-1][j-1] + 1);
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}
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}
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}
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return dp[word1.length][word2.length];
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};
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```
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-----------------------
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* 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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* B站视频：[代码随想录](https://space.bilibili.com/525438321)
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* 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码.jpg width=450> </img></div>
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