三种操作的转移:

    1. class Solution {
    2. public:
    3. int minDistance(string word1, string word2) {
    4. int n = word1.size(), m = word2.size();
    5. vector<vector<int>> dp(n + 1, vector<int>(m + 1, 0));
    6. for(int i = 1; i <= m; i++)
    7. dp[0][i] = i;
    8. for(int i = 1; i <= n; i++)
    9. dp[i][0] = i;
    10. for(int i = 1; i <= n; i++)
    11. {
    12. for(int j = 1; j <= m; j++)
    13. {
    14. 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
    15. dp[i][j] = min(dp[i][j], dp[i - 1][j] + 1);//删掉一个
    16. dp[i][j] = min(dp[i][j], dp[i][j - 1] + 1);//添加一个
    17. }
    18. }
    19. return dp[n][m];
    20. }
    21. };

    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()];
        }
    };