406 lines
12 KiB
Markdown
406 lines
12 KiB
Markdown
<p align="center">
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<a href="https://programmercarl.com/other/xunlianying.html" target="_blank">
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<img src="../pics/训练营.png" width="1000"/>
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</a>
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<p align="center"><strong><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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[力扣题目链接](https://leetcode.cn/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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### 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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有同学问了，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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为什么这么定义我在 [718. 最长重复子数组](https://programmercarl.com/0718.最长重复子数组.html) 中做了详细的讲解。
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其实用i来表示也可以！ 用i-1就是为了方便后面dp数组初始化的。
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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删除一个元素，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 再加上一个操作。
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即 `dp[i][j] = dp[i - 1][j] + 1;`
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* 操作二：word2删除一个元素，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 再加上一个操作。
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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，怎么都是删除元素，添加元素去哪了。
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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'` 和 `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，此时不用增删加元素。
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可以回顾一下，`if (word1[i - 1] == word2[j - 1])`的时候我们的操作 是 `dp[i][j] = dp[i - 1][j - 1]` 对吧。
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那么只需要一次替换的操作，就可以让 word1[i - 1] 和 word2[j - 1] 相同。
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所以 `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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### 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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### 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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* 时间复杂度: O(n * m)
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* 空间复杂度: O(n * m)
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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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TypeScript：
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```typescript
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function minDistance(word1: string, word2: string): number {
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/**
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dp[i][j]: word1前i个字符，word2前j个字符，最少操作数
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dp[0][0]=0：表示word1前0个字符为'', word2前0个字符为''
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*/
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const length1: number = word1.length,
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length2: number = word2.length;
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const dp: number[][] = new Array(length1 + 1).fill(0)
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.map(_ => new Array(length2 + 1).fill(0));
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for (let i = 0; i <= length1; i++) {
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dp[i][0] = i;
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}
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for (let i = 0; i <= length2; i++) {
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dp[0][i] = i;
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}
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for (let i = 1; i <= length1; i++) {
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for (let j = 1; j <= length2; 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(
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dp[i - 1][j],
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dp[i][j - 1],
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dp[i - 1][j - 1]
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) + 1;
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}
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}
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}
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return dp[length1][length2];
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};
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```
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C：
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```c
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int min(int num1, int num2, int num3) {
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return num1 > num2 ? (num2 > num3 ? num3 : num2) : (num1 > num3 ? num3 : num1);
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}
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int minDistance(char * word1, char * word2){
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int dp[strlen(word1)+1][strlen(word2)+1];
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dp[0][0] = 0;
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for (int i = 1; i <= strlen(word1); i++) dp[i][0] = i;
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for (int i = 1; i <= strlen(word2); i++) dp[0][i] = i;
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for (int i = 1; i <= strlen(word1); i++) {
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for (int j = 1; j <= strlen(word2); 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][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[strlen(word1)][strlen(word2)];
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}
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```
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<p align="center">
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<a href="https://programmercarl.com/other/kstar.html" target="_blank">
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<img src="../pics/网站星球宣传海报.jpg" width="1000"/>
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</a>
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