给定三个字符串 s1、s2、s3,请你帮忙验证 s3 是否是由 s1 和 s2 交错 组成的。
两个字符串 s 和 t 交错 的定义与过程如下,其中每个字符串都会被分割成若干 非空 子字符串:
s = s1 + s2 + … + sn
t = t1 + t2 + … + tm
|n - m| <= 1
交错 是 s1 + t1 + s2 + t2 + s3 + t3 + … 或者 t1 + s1 + t2 + s2 + t3 + s3 + …
注意:a + b 意味着字符串 a 和 b 连接。
示例 1:
输入:s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbcbcac”
输出:true
示例 2:
输入:s1 = “aabcc”, s2 = “dbbca”, s3 = “aadbbbaccc”
输出:false
示例 3:
输入:s1 = “”, s2 = “”, s3 = “”
输出:true
提示:
0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1、s2、和 s3 都由小写英文字母组成
题目链接
class Solution:def isInterleave(self, s1: str, s2: str, s3: str) -> bool:len1, len2 = len(s1), len(s2)if len(s3) != len1 + len2:return False## dp[i][j] 表示s1 前 i 个字符和 s2 前 j 个字符能组成交错字符串dp = [[0]*(len2+1) for i in range(len1+1)] dp[0][0] = 1for i in range(1, len1+1):if s1[i-1] == s3[i-1]:dp[i][0] = 1else:breakfor i in range(1, len2+1):if s2[i-1] == s3[i-1]:dp[0][i] = 1else:breakfor i in range(1,len1+1):for j in range(1, len2+1):if dp[i-1][j] == 1 and s1[i-1] == s3[i+j-1]:dp[i][j] = 1if dp[i][j-1] == 1 and s2[j-1] == s3[i+j-1]:dp[i][j] = 1 if dp[len1][len2] == 1:return Trueelse:return False