题目描述

给定两个单词 word1 和 word2，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')

我的解法——DP（AC）

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        if(not len(word1) or not len(word2)):
            return abs(len(word1) - len(word2))
        
        dp = []
        for i in range(len(word1) + 1):
            dp.append([0] * (len(word2) + 1))
        
        for i in range(1, len(word1) + 1):
            dp[i][0] = dp[i - 1][0] +1
        for i in range(1, len(word2) + 1):
            dp[0][i] = dp[0][i - 1] + 1

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