代码随想录算法训练营第55天|● 583. 两个字符串的删除操作 ● 72. 编辑距离 ● 编辑距离总结篇

Yueqiao132024-05-31 16:37

就是两个都要删除，取两个都删的最低操作

python 复制代码

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:
                    dp[i][j]=min(dp[i-1][j-1]+2,dp[i-1][j]+1,dp[i][j-1]+1)
        return dp[-1][-1]

72. 编辑距离

增删替--增==对面删删==删替：+1

python 复制代码

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

编辑距离正式结束