72.编辑距离

三种操作的转移：

class Solution {
public:
    int minDistance(string word1, string word2) {
        int n = word1.size(), m = word2.size();
        vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
        for(int i = 1; i <= m; i++)
            dp[0][i] = i;
        for(int i = 1; i <= n; i++)
            dp[i][0] = i;
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
            {
                dp[i][j] = dp[i - 1][j - 1] + (word1[i - 1] != word2[j -1]);//替换一个 这里word1 word2的是index所以减1
                dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);//删掉一个
                dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1);//添加一个
            }
        }
        return dp[n][m];
    }
};

第二次写题

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector<vector<int>> distance = vector<vector<int>>(word1.size() + 1, vector<int>(word2.size() + 1, INT_MIN));

        for(int i = 0; i <= word1.size(); i++)
            distance[i][0] = i;
        for(int i = 0; i <= word2.size(); i++)
            distance[0][i] = i;
        for(int i = 1; i <= word1.size(); i++)
        {
            for(int j = 1; j <= word2.size(); j++)
            {
                distance[i][j] = distance[i - 1][j - 1] + (word1[i - 1] != word2[j - 1]); //替换；
                distance[i][j] = min(distance[i][j], distance[i - 1][j] + 1);//插入
                distance[i][j] = min(distance[i][j], distance[i][j - 1] + 1);//删除
            }
        }
        return distance[word1.size()][word2.size()];
    }
};

第二次写题

class Solution {
public:
    int minDistance(string word1, string word2) {
        vector<vector<int>> distance = vector<vector<int>>(word1.size() + 1, vector<int>(word2.size() + 1, INT_MIN));

        for(int i = 0; i <= word1.size(); i++)
            distance[i][0] = i;
        for(int i = 0; i <= word2.size(); i++)
            distance[0][i] = i;
        for(int i = 1; i <= word1.size(); i++)
        {
            for(int j = 1; j <= word2.size(); j++)
            {
                distance[i][j] = distance[i - 1][j - 1] + (word1[i - 1] != word2[j - 1]); //替换；
                distance[i][j] = min(distance[i][j], distance[i - 1][j] + 1);//插入
                distance[i][j] = min(distance[i][j], distance[i][j - 1] + 1);//删除
            }
        }
        return distance[word1.size()][word2.size()];
    }
};