题目来源：
https://leetcode.com/problems/interleaving-string/
题意分析：
给定字符串s1，s2,s3，判断s3是否由s1和s2穿插组成。如“abc”由“ac”，“b”组成，而“cba”不是。
题目思路：
这是一个动态规划问题。令ans[i][j]为s1[:i]和s2[:j]匹配是否成功。那么动态方程是if s1[i - 1] == s3[i + j - 1] 那么ans[i][j] = ans[i][j] or ans[i - 1][j];if s2[j - 1] == s3[i + j - 1] 那么ans[i][j] = ans[i][j] or ans[i][j - 1]。
代码（python)：
class Solution(object):
def isInterleave(self, s1, s2, s3):
"""
:type s1: str
:type s2: str
:type s3: str
:rtype: bool
"""
i,j,k = 0,0,0
m,n,t = len(s1),len(s2),len(s3)
if m + n != t:
return False
ans = [[False for i in range(n+1)] for j in range(m+1)]
ans[0][0] = True
for i in range(1,m+1):
if s1[i - 1] == s3[i - 1]:
ans[i][0] = True
else:
break
for i in range(1,n + 1):
if s2[i - 1] == s3[i -1]:
ans[0][i] = True
else:
break
for i in range(1,m + 1):
for j in range(1,n + 1):
if s1[i - 1] == s3[i + j - 1]:
ans[i][j] = ans[i][j] or ans[i - 1][j]
if s2[j - 1] == s3[i + j - 1]:
ans[i][j] = ans[i][j] or ans[i][j - 1]
return ans[m][n]