题目如下：

解题思路：本题可以采用动态规划的方法。记dp[i][j]表示用s1的前i个字符和s2的前j个字符能否组成s3的前(i+j)个字符，如果dp[i-1][j]是True的话，只要s1[i] == s3[i-1+j]，那么dp[i][j]就为True；同理，如果dp[i][j-1]是True，只要满足s2[j] == s3[i-1+j]，那么dp[i][j]也为True。

代码如下：

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) == 0:            return len(s3) == 0        elif len(s1) == 0:            return s2 == s3        elif len(s2) == 0:            return s1 == s3        elif len(s1) + len(s2) != len(s3):            return False        dp = [ [0]* (len(s2) + 1)  for i in range(len(s1)+1) ]        if s1[0] == s3[0]:            dp[1][0] = 1        if s2[0] == s3[0]:            dp[0][1] = 1        for i in range(len(dp)):            for j in range(len(dp[i])):                if i > 0 and j > 0 and dp[i-1][j] == 0 and dp[i][j-1] == 0:                    dp[i][j] = 0                elif i > 0 and dp[i-1][j] == 1 and s1[i-1] == s3[i-1+j]:                    dp[i][j] = 1                elif j > 0 and dp[i][j-1] == 1 and s2[j-1] == s3[i+j-1]:                    dp[i][j] = 1        #print dp        return dp[-1][-1] == 1