leetcode刷题python之编辑距离

思路：使用动态规划，dp[i][j]表示，将word1[:j]变成word2[:i]需要的步骤数。如果word1[j - 1] == word2[i - 1]，表示有相同字符出现，只需要进行这个字符前面的匹配就可以dp[i][j] = dp[i - 1][j - 1]。如果word1[j - 1] ！= word2[i - 1]，则在步数最少的前一次基础上进行加一，即dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1。

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        n1 = len(word1)
        n2 = len(word2)
        dp = [[0 for _ in range(n1 + 1)] for _ in range(n2 + 1)]
        for i in range(n1 + 1):
            dp[0][i] = i
        for i in range(n2 + 1):
            dp[i][0] = i
        for i in range(1, n2 + 1):
            for j in range(1, n1 + 1):
                if word1[j - 1] == word2[i - 1]:
                    dp[i][j] = dp[i - 1][j - 1]
                else:
                    dp[i][j] = min(dp[i - 1][j - 1], dp[i - 1][j], dp[i][j - 1]) + 1