[Leetcode]@python 97. Interleaving String

题目链接

https://leetcode.com/problems/interleaving-string/

题目原文

Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2.

For example,
Given:
s1 = "aabcc",
s2 = "dbbca",

When s3 = "aadbbcbcac", return true.
When s3 = "aadbbbaccc", return false.

题目大意

给定字符串s1, s2, s3,判断s3是否由s1和s2交叉组成的

解题思路

动态规划：dp[i][j]表示s1[0...i-1]和s2[0...j-1]是否可以拼接为s3[0...i+j-1]，可以拼接为true，不可以拼接为false

代码

class Solution(object):
    def isInterleave(self, s1, s2, s3):
        """
        :type s1: str
        :type s2: str
        :type s3: str
        :rtype: bool
        """
        if len(s1) + len(s2) != len(s3):
            return False
        dp = [[False for i in range(len(s2) + 1)] for j in range(len(s1) + 1)]
        dp[0][0] = True

        for i in range(1, len(s1) + 1):
            if dp[i - 1][0] and s3[i - 1] == s1[i - 1]:
                dp[i][0] = True
        for i in range(1, len(s2) + 1):
            if dp[0][i - 1] and s3[i - 1] == s2[i - 1]:
                dp[0][i] = True

        for i in range(1, len(s1) + 1):
            for j in range(1, len(s2) + 1):
                if (dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]) or (dp[i][j - 1] and s2[j - 1] == s3[i + j - 1]):
                    dp[i][j] = True
        return dp[len(s1)][len(s2)]