leetcode Interleaving String python 解法

標籤：

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.

 

方案1：DFS

class Solution:    # @return a boolean    def isInterleave(self, s1, s2, s3):        if len(s1)+len(s2) == len(s3):            return self.DFS(s1,s2,s3)        else:            return False                def DFS(self,s1,s2,s3):        if not (s1 and s2) :            if s3 == s1+s2:                return True            else:                return False                    if s1[0] == s3[0] and self.DFS(s1[1:],s2,s3[1:]):            return True        elif s2[0] == s3[0] and self.DFS(s1,s2[1:],s3[1:]):            return True        else:            return False    

很不幸的是，逾時了

方案2：BFS

import Queueclass Solution:    # @return a boolean    def isInterleave(self, s1, s2, s3):        if len(s1)+len(s2) == len(s3):            return self.BFS(s1,s2,s3)        else:            return False                def BFS(self,s1,s2,s3):        q = Queue.Queue(len(s3))        q.put((0,0))        while(not q.empty()):            (x,y) = q.get()            if x==len(s1) or y == len(s2):                if s3[x+y:] == s1[x:]+s2[y:]:                    return True                else:                    return False            if s1[x] == s3[x+y]:                q.put((x+1,y))            if s2[y] == s3[x+y]:                q.put((x,y+1))                return False

還是逾時

leetcode Interleaving String python 解法