583. 两个字符串的删除操作

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        m, n = len(word1), len(word2)
        dp = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(1, m + 1):
            dp[i][0] = i
        for j in range(1, n + 1):
            dp[0][j] = j
        for i in range(1, m + 1):
            for j in range(1, n + 1):
                if word1[i - 1] == word2[j - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    dp[i][j] = 1 + min(dp[i - 1][j], dp[i][j - 1])
        return dp[-1][-1]

class Solution {
    public int minDistance(String word1, String word2) {
        int m = word1.length(), n = word2.length();
        int[][] dp = new int[m + 1][n + 1];
        for (int i = 1; i <= m; ++i) {
            dp[i][0] = i;
        }
        for (int j = 1; j <= n; ++j) {
            dp[0][j] = j;
        }
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                } else {
                    dp[i][j] = 1 + Math.min(dp[i - 1][j], dp[i][j - 1]);
                }
            }
        }
        return dp[m][n];
    }
}

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));
        for (int i = 1; i <= m; ++i) dp[i][0] = i;
        for (int j = 1; 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] = 1 + min(dp[i - 1][j], dp[i][j - 1]);
            }
        }
        return dp[m][n];
    }
};

func minDistance(word1 string, word2 string) int {
	m, n := len(word1), len(word2)
	dp := make([][]int, m+1)
	for i := range dp {
		dp[i] = make([]int, n+1)
		dp[i][0] = i
	}
	for j := range dp[0] {
		dp[0][j] = j
	}
	for i := 1; i <= m; i++ {
		for j := 1; j <= n; j++ {
			if word1[i-1] == word2[j-1] {
				dp[i][j] = dp[i-1][j-1]
			} else {
				dp[i][j] = 1 + min(dp[i-1][j], dp[i][j-1])
			}
		}
	}
	return dp[m][n]
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}