【leetcode】Edit Distance（编辑距离）（动态规划）

1、题目描述

Given two words word1 and word2, find the minimum number of operations required to convert word1 to word2.

You have the following 3 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')

2、问题描述：

求两个字符串如何变成一样的，最少的步骤，典型的动态规划问题。

3、问题关键：

对于当前比较的两个字符 word1[i] 和 word2[j]，若二者相同，一切好说，直接跳到下一个位置。若不相同，有三种处理方法，首先是直接插入一个 word2[j]，那么 word2[j] 位置的字符就跳过了，接着比较 word1[i] 和 word2[j+1] 即可。第二个种方法是删除，即将 word1[i] 字符直接删掉，接着比较 word1[i+1] 和 word2[j] 即可。第三种则是将 word1[i] 修改为 word2[j]，接着比较 word1[i+1] 和 word[j+1] 即可。
状态表示：f[i][j] 表示word1的前i个字符变成word2的前j个字符，最少需要操作数。
状态转移：四种情况：

1.如果word1[i] == word2[j],则其操作数为f[i-1][j - 1];
2.将word1[i]修改word2[j]为一样的，则操作数为：f[i - 1][j -1] + 1;
3.将word1[i] 删除，则其操作次数等于f[i - 1][j] + 1;
4.将word1[i]前word2[j],则其i操作数等于f[i][j - 1] + 1;

时间复杂度 $O(n^2)$ .

4、C++代码：

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));
        for (int i = 0; i <= m; ++i) dp[i][0] = i;
        for (int i = 0; i <= n; ++i) dp[0][i] = i;
        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], min(dp[i - 1][j], dp[i][j - 1])) + 1;
                }
            }
        }
        return dp[m][n];
    }
};