题干：

代码：

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector<vector<int>>dp(word1.size() + 1, vector<int>(word2.size() + 1));
        for(int i = 0; i <= word1.size(); i++)dp[i][0] = i;
        for(int j = 0; j <= word2.size(); j++)dp[0][j] = j;

        for(int i = 1; i <= word1.size(); i++){
            for(int j = 1; j <= word2.size(); 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], dp[i][j - 1], dp[i - 1][j - 1]}) + 1;
            }
        }
        return dp[word1.size()][word2.size()];
    }
};

1.定义：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]。

2.递推公式：

if (word1[i - 1] == word2[j - 1])
    不操作
if (word1[i - 1] != word2[j - 1])
    增
    删
    换