LeetCode解题报告（428）-- 72. Edit Distance

Problem

Given two strings word1 and word2, return the minimum number of operations required to convert word1 to word2.

You have the following three operations permitted on a word:

• Insert a character
• Delete a character
• Replace a character

Example 1:

Input: word1 = "horse", word2 = "ros"
Output: 3
Explanation: 
horse -> rorse (replace 'h' with 'r')
rorse -> rose (remove 'r')
rose -> ros (remove 'e')

Example 2:

Input: word1 = "intention", word2 = "execution"
Output: 5
Explanation: 
intention -> inention (remove 't')
inention -> enention (replace 'i' with 'e')
enention -> exention (replace 'n' with 'x')
exention -> exection (replace 'n' with 'c')
exection -> execution (insert 'u')

Constraints:

• 0 <= word1.length, word2.length <= 500
• word1 and word2 consist of lowercase English letters.

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Analysis

  这是一道非常非常经典的动态规划题目，也是一个在工业实践中很常用的算法。首先我们来看什么是编辑距离。给定两个字符串word1和word2，通过一系列的操作把word1转化为word2，操作包括在word1中插入、删除、替换。这种两个字符串变换的题目本身就很适合二维dp。dp[i][j]表示把word1前i个字符转换成word2前j个字符所需要的最小编辑距离。

  先来看base case。当word2长度为0时，要把word1变过去只能是删除所有的字符，所以dp[i][0] = i。同理，如果word1长度为0时，要把word1变成word2只能是添加字符，所以dp[0][j] = j。

  然后来看转移方程。当word1[i - 1] == word2[j - 1]时，不需要编辑操作，所以dp[i][j] = dp[i - 1][j - 1]；当word1[i - 1] != word2[j - 1]时，有三种操作：

• word1[i - 1]替换成word2[j - 1]，所以dp[i][j] = dp[i - 1][j - 1]；
• 删除掉word1[i - 1]，用下一个继续和word2[j - 1]匹配，所以dp[i][j] = dp[i - 1][j]；
• 在word1中增加word2[j - 1]，所以dp[i][j] = dp[i][j - 1].

Solution

  无

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Code

class Solution {
public:
    int minDistance(string word1, string word2) {
        int m = word1.size(), n = word2.size();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        
        for (int i = 0; i <= m; i++) {
            dp[i][0] = i;
        }
        for (int j = 0; j <= n; j++) {
            dp[0][j] = j;
        }
        
        for (int i = 1; i <= m; i++) {
            for (int j = 1; j <= n; j++) {
                if (word1[i - 1] == word2[j - 1]) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = min({dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]}) + 1;
                }
            }
        }
        return dp[m][n];
    }
};

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Summary

  这道题目是非常经典的dp，需要熟练掌握。它是很多字符串转换dp的原型，很多变形都是基于这道题目出的。这道题目的分享到这里，感谢你的支持！