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]的字符串区间是否能组成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()];    }};