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  • 【leetcode】583. Delete Operation for Two Strings

    题目如下:

    解题思路:一般这种求最大/最小的题目大多数都是用动态规划。记dp[i][j] = n 表示最少经过n次删除操作后,使得word1[0~i]与word2[0~j]相等。那么可以等到递推关系式,如果word1[i] == word2[j],那么dp[i][j] = dp[i-1][j-1];如果不相等,要么删除word1[i],要么删除word2[j],那么dp[i][j] = min(dp[i-1][j],dp[i][j-1])+1。

    代码如下:

    class Solution(object):
        def minDistance(self, word1, word2):
            """
            :type word1: str
            :type word2: str
            :rtype: int
            """
            if len(word1) == 0 or len(word2) == 0:
                return abs(len(word2) - len(word1))
            dp = [[0 for x in range(len(word2)+1)] for x in range(len(word1)+1)]
            for i in xrange(1,len(word1)+1):
    #如果word2是空,表示应该删除word1里面全部的字符,所以记dp[i][0] = i dp[i][0]
    = i for j in xrange(1,len(word2)+1): dp[0][j] = j for i in xrange(1,len(word1)+1): for j in xrange(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],dp[i][j-1])+1 return dp[-1][-1]
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  • 原文地址:https://www.cnblogs.com/seyjs/p/9258240.html
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