代码随想录day56|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:
        # dp = [[0]*(len(word2)+1) for _ in range(len(word1)+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] = max(dp[i-1][j],dp[i][j-1])
        # return len(word1)+len(word2)-2*dp[len(word1)][len(word2)] 
        dp = [[0]*(len(word2)+1) for _ in range(len(word1)+1)]
        #dp数组的含义是最小删除的步数
        for i in range(len(word1)+1):
            dp[i][0] = i
        for j in range(len(word2)+1):
            dp[0][j] = j
        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]
                else:
                    dp[i][j] = min(dp[i-1][j]+1,dp[i][j-1]+1,dp[i-1][j-1]+2)
        return dp[len(word1)][len(word2)] 

72. 编辑距离

class Solution:
    def minDistance(self, word1: str, word2: str) -> int:
        dp = [[0] *(len(word2)+1) for _ in range(len(word1)+1)]
        for i in range(len(word1)+1):
            dp[i][0] = i
        for j in range(len(word2)+1):
            dp[0][j] = j 
        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]
                else:
                    #用增删改三个操作，对word1删相当于对word2增，所以增删操作保留一个就可以
                    dp[i][j]=min(dp[i][j-1]+1,dp[i-1][j]+1,dp[i-1][j-1]+1)
        return dp[len(word1)][len(word2)]