LeetCode_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.

方法一： DFS，大数据超时

class Solution {
public:
    bool dfs(int i1, int i2, int i3, string &s1, string &s2, string &s3)
    {

       if(i1 == size1 && i2 == size2  ) return true;

       if(i1 == size1){
           if(s2[i2] == s3[i3])
               return  dfs(i1,i2+, i3+, s1, s2, s3);
            else
              return false;

       }else if(i2 == size2){
           if(s1[i1] == s3[i3])
                  return  dfs(i1+, i2, i3+, s1, s2, s3);
             else
            return false ;

       }else{
            bool f1, f2;
            if(s1[i1] == s3[i3])
            {

                f1 =  dfs(i1+, i2, i3+, s1, s2, s3);
                if(f1 == true)   return true ;
            }
            if(s2[i2] == s3[i3])
            {
               f2 = dfs(i1,i2+, i3+, s1, s2, s3);

               return f2;
            }
            return false;

       }

    }
    bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        size1 = s1.size();
        size2 = s2.size();
        size3 = s3.size();
        if(size1 + size2 != size3 ) return false;

      return    dfs(,,, s1, s2, s3) ;

    }
private:
    int size1;
    int size2;
    int size3;
};

方法二： 2D 动态规划

Use match[i][j] to save if s1[i] or s2[j] are matched s3[i+j].

match[i][j] =   (s3[i+j]==s1[i]  && match[i-1][j])
                    || (s3[i+j] ==s2[j] && match[i][j-1])

class Solution {
public:

    bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        size1 = s1.size();
        size2 = s2.size();
        size3 = s3.size();
        if(size1 + size2 != size3 ) return false;

        vector<vector<bool>> match( size1+, vector<bool>(size2+)) ;

        match[][] = true;

        for(int i = ; i <= size1 ; i++)
        {
         if(s1[i-] == s3[i-])
           match[i][] = true;
          else
              break;
        }

        for(int j = ; j <= size2 ; j++)
        {
            if(s2[j-] == s3[j-])
               match[][j] = true;
             else
               break;
        }

        int i,j;
        for(i = ;  i <= size1 ; i++)
          for(j = ; j <= size2; j++)
          {
              match[i][j] =( match[i-][j]&&s1[i-] == s3[i+j-]) ||
                             (match[i][j-] && s2[j-] == s3[i+j-]) ;
          }

        return    match[size1][size2] ;

    }
private:
    int size1;
    int size2;
    int size3;
};