583. Delete Operation for Two Strings

https://leetcode.com/problems/delete-operation-for-two-strings/description/

对最长公共子序列的计算分析见：https://blog.csdn.net/hrn1216/article/details/51534607

Description

Given two words word1 and word2, find the minimum number of steps required to make word1 and word2 the same, where in each step you can delete one character in either string.

Example 1:

Input: "sea", "eat"
Output: 2
Explanation: You need one step to make "sea" to "ea" and another step to make "eat" to "ea".

Note:

The length of given words won't exceed 500.
Characters in given words can only be lower-case letters.

Solution

Solution-1

用递归的动态规划（DP）的方式求解：

class Solution(object):
    def LCS(self, len1, len2):
        if len1==0 or len2==0: 
            return 0
        if (len1, len2) in self.record:
            return self.record[(len1, len2)]
        if self.word1[len1-1]==self.word2[len2-1]:
            ret = self.LCS(len1-1, len2-1)+1
            self.record[(len1, len2)] = ret
        else:
            ret = max(self.LCS(len1-1, len2),
                     self.LCS(len1, len2-1))
            self.record[(len1, len2)] = ret
        return ret
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        self.record = dict()
        self.word1, len1 = word1, len(word1)
        self.word2, len2 = word2, len(word2)
        return len1+len2-2*self.LCS(len1, len2)

Solution-2

非递归方式：

class Solution(object):
    def minDistance(self, word1, word2):
        """
        :type word1: str
        :type word2: str
        :rtype: int
        """
        len1, len2 = len(word1), len(word2)
        record = [[0]*(len2+1) for i in range(len1+1)]
        for i in range(1, len1+1):
            for j in range(1, len2+1):
                if word1[i-1]==word2[j-1]: 
                    record[i][j] = record[i-1][j-1]+1
                else:
                    record[i][j] = max(record[i-1][j], record[i][j-1])
        return len1+len2-2*record[len1][len2]