【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]

 

转载于:https://www.cnblogs.com/seyjs/p/9258240.html