原题地址：https://oj.leetcode.com/problems/interleaving-string/
题意：
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.
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: # @return a boolean def isInterleave(self, s1, s2, s3): 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): dp[i][0] = dp[i-1][0] and s3[i-1]==s1[i-1] for i in range(1,len(s2)+1): dp[0][i] = dp[0][i-1] and s3[i-1]==s2[i-1] for i in range(1,len(s1)+1): for j in range(1,len(s2)+1): dp[i][j] = (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]) return dp[len(s1)][len(s2)]