leetcode_question_97 Interleaving String

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.

Recurse:
Judge Small: Accepted! 
Judge Large: Time Limit Exceeded

bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        if(s1.length() == 0) return s3 == s2;
    	if(s2.length() == 0) return s3 == s1;
		if(s3.length() == 0) return s1.length() + s2.length() == 0;
		
		if(s1[0] == s3[0] && s2[0] != s3[0])
			return isInterleave(s1.substr(1), s2, s3.substr(1));
		else if(s1[0] != s3[0] && s2[0] == s3[0])
			return isInterleave(s1, s2.substr(1), s3.substr(1));
		else if(s1[0] == s3[0] && s1[0] == s3[0])
			return isInterleave(s1.substr(1), s2, s3.substr(1)) || isInterleave(s1, s2.substr(1), s3.substr(1));
		else
			return false;
    }

2-dimension dp:

这是一个二维的动态规划，

s1 = "aabcc"
s2 = "dbbca"
s3 = "aadbbcbcac"

class Solution {
public:
    bool dp[101][101];
    bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
    	int size1 = s1.length();
		int size2 = s2.length();
		int size3 = s3.length();
		if( size1 + size2 != size3) return false;
		
		memset(dp, false, sizeof(bool)*101*101);
		dp[0][0] = true;
        for(int i = 1; i <= size1; ++i)
			if(s1[i-1] == s3[i-1]) dp[i][0] = true;
			else break;
		for(int j = 1; j <= size2; ++j)
			if(s2[j-1] == s3[j-1]) dp[0][j] = true;
			else break;

		int k;
		for(int i = 1; i <= size1; ++i)
			for(int j = 1; j <= size2; ++j)
			{
				k = i + j;
				if(s1[i-1] == s3[k-1]) dp[i][j] = dp[i-1][j] || dp[i][j];
				if(s2[j-1] == s3[k-1]) dp[i][j] = dp[i][j-1] || dp[i][j];
			}
		return dp[size1][size2];
    }
};