算法刷题Day 56两个字符串的删除操作+编辑距离

Day 56 动态规划

583. 两个字符串的删除操作

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 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][j - 1] + 1, dp[i - 1][j - 1] + 2));
                }
            }
        }

        return dp[m][n];
    }
};

72. 编辑距离

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

        return dp[m][n];
    }
};