动态规划解法,这里的dp[i][j]表示word1[0----i]变化到word2[0----j]所需要的最小操作步骤。

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],min(dp[i-1][j],dp[i][j-1])) + 1;
            }
        }
        
        return dp[m][n];
    }
};