LeetCode 72. 编辑距离

给你两个单词?word1 和?word2，请你计算出将?word1?转换成?word2 所使用的最少操作数?。

你可以对一个单词进行如下三种操作：

插入一个字符
删除一个字符
替换一个字符

示例?1：

输入：word1 = "horse", word2 = "ros"
输出：3
解释：
horse -> rorse (将 ‘h‘ 替换为 ‘r‘)
rorse -> rose (删除 ‘r‘)
rose -> ros (删除 ‘e‘)
示例?2：

输入：word1 = "intention", word2 = "execution"
输出：5
解释：
intention -> inention (删除 ‘t‘)
inention -> enention (将 ‘i‘ 替换为 ‘e‘)
enention -> exention (将 ‘n‘ 替换为 ‘x‘)
exention -> exection (将 ‘n‘ 替换为 ‘c‘)
exection -> execution (插入 ‘u‘)

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

LeetCode 72. 编辑距离

原文：https://www.cnblogs.com/sandy-t/p/13287279.html