代码随想训练营第五十六天（Python）｜ 583. 两个字符串的删除操作、72. 编辑距离

583. 两个字符串的删除操作

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        n, m = len(word1), len(word2)
        # dp 数组代表使得 word1 以 i-1 结尾和 word2 以 j-1 结尾相同的最小步数
        dp = [[0] * (m+1) for _ in range(n+1)]

        # 初始化
        for i in range(n+1):
            dp[i][0] = i

        for j in range(m+1):
            dp[0][j] = j

        for i in range(1, n+1):
            for j in range(1, m+1):
                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, dp[i][j-1] + 1, dp[i-1][j-1] + 2)

        return dp[n][m]

72. 编辑距离

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        n, m = len(word1), len(word2)
        # dp 数组代表以 i-1 结尾的 word1 变成 以 j-1 结尾的 word2 的最小操作数为 dp[i][j]
        dp = [[0] * (m+1) for _ in range(n+1)]
        # 初始化
        for i in range(n+1):
            dp[i][0] = i

        for j in range(m+1):
            dp[0][j] = j

        for i in range(1, n+1):
            for j in range(1, m+1):
                if word1[i-1] == word2[j-1]:
                    dp[i][j] = dp[i-1][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j], dp[i][j-1], dp[i-1][j-1]) + 1

        return dp[n][m]