LeetCode 72. 编辑距离 Python

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：

1. 插入一个字符
2. 删除一个字符
3. 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')

代码：

class Solution:
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        n = len(word1)
        m = len(word2)

        dp = [[0]*(m+1) for x in range(n+1)]     #dp[i][j]是word1的长度i位置替换world2的长度j位置所需要的编辑距离
        
        for i in range(0,n+1):
            for j in range(0,m+1):
                if i == 0:
                    dp[0][j] = j
                elif j ==0:
                    dp[i][0] = i
                else:
                    dp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1)
                    if word1[i-1] == word2[j-1]:
                        dp[i][j] = min(dp[i-1][j-1],dp[i][j])

                    else:
                        dp[i][j] = min(dp[i-1][j-1] + 1,dp[i][j])

        return dp[n][m]