[Leetcode]@python 97. Interleaving String-优快云博客

本文针对LeetCode上的一道经典题目——交错字符串问题进行解析。通过动态规划的方法来判断一个字符串是否能由另外两个字符串交错组成。文章详细介绍了解题思路，并提供了一段Python实现的代码。

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)]

转载于:https://www.cnblogs.com/slurm/p/5221580.html