Interleaving String——是否由两个string交叉、DP

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() == ) return s3 == s2;
         if(s2.length() == ) return s3 == s1;
         if(s3.length() == ) return s1.length() + s2.length() == ;  

         if(s1[] == s3[] && s2[] != s3[])
             return isInterleave(s1.substr(), s2, s3.substr());
         else if(s1[] != s3[] && s2[] == s3[])
             return isInterleave(s1, s2.substr(), s3.substr());
         else if(s1[] == s3[] && s1[] == s3[])
             return isInterleave(s1.substr(), s2, s3.substr()) || isInterleave(s1, s2.substr(), s3.substr());
         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 isInterleave(string s1, string s2, string s3) {
         if(s3.length()!=s1.length()+s2.length()) return false;
         vector<vector<bool>> res(s1.length()+,vector<bool>(s2.length()+, false));
         res[][]=true;
         for(int i=;i<=s1.length();i++){
             res[i][]=res[i-][]&&s1[i-]==s3[i-];
         }
         for(int j=;j<=s2.length();j++){
             res[][j]=res[][j-]&&s2[j-]==s3[j-];
         }
         for(int i=;i<=s1.length();i++){
             for(int j=;j<=s2.length();j++){
                 res[i][j]=(res[i-][j]&&s1[i-]==s3[i+j-])||(res[i][j-]&&s2[j-]==s3[i+j-]);
             }
         }
         return res[s1.length()][s2.length()];
     }

 };