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.

//利用dp解决，dp[i][j]的状态表示为s1[0...i] + s2[0...j]的字符串区间能否组成s3.
//那么，动态转移方程为:
// 1） s1[i-1] == s3[i+j-1] && dp[i-1][j] = true 那么，dp[i][j] = true;
// 2） s2[j-1] == s3[i+j-1] && dp[i][j-1] = true 那么，dp[i][j] = true;
class Solution {
public:
    bool isInterleave(std::string s1, std::string s2, std::string s3) {
		if(s1.size() + s2.size() != s3.size()) return false;
		std::vector
  
   > dp(s1.size()+1,std::vector
   
    (s2.size()+1,0)); dp[0][0] = 1; for (int i = 1; i < s1.size()+1; i++) { if(s1[i-1] == s3[i-1] && dp[i-1][0]) dp[i][0] = true; } for (int i = 1; i < s2.size()+1; i++) { if(s2[i-1] == s3[i-1] && dp[0][i-1]) dp[0][i] = true; } for (int i = 1; i < s1.size() + 1; i++) { for (int j = 1; j < s2.size() + 1; j++) { if(s1[i-1] == s3[i+j-1] && dp[i-1][j]) dp[i][j] = true; if(s2[j-1] == s3[i+j-1] && dp[i][j-1]) dp[i][j] = true; } } return dp[s1.size()][s2.size()]; } };