[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.

动规题。用A[i1][i2]表示s1中的前i1个元素和s2中前i2个元素能否构成s3的前i1+i3个元素。

A[i1][i2] = (A[i1 - 1][i2] && s1[i1] == s3[i1+i2]) || (A[i1][i2 - 1] && s2[i2] == s3[i1+i2])

class Solution {
public:
    int **mark;
    bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(s1.length() + s2.length() != s3.length())
            return false;
        if(s1.length() == 0 && s2.length() == 0)
            return true;
        if(s1.length() == 0)
            return s2 == s3;
        if(s2.length() == 0)
            return s1 == s3;
        mark = new int*[s1.length()+1];
        for(int i = 0;i<s1.length()+1;i++)
        {
            mark[i] = new int[s2.length()+1];
            for(int j = 0;j < s2.length()+1;j++)
                mark[i][j] = -1;
        }
        return check(&s1, &s2, &s3,0,0,0);
    }
    bool check(string *s1, string *s2, string *s3,int i1,int i2,int i3)
    {
        bool tag1;
        if(mark[i1][i2] != -1)
        {
            return bool(mark[i1][i2]);
        }
        if(i3 == s3->length())
        {
            mark[i1][i2] = 1;
            return true;
        }
        if(i1 < s1->length())
        {
            if(s3->at(i3) == s1->at(i1))
            {
                tag1 = check(s1,s2,s3,i1+1,i2,i3+1);
                if(tag1 == true)
                {
                    mark[i1][i2] = 1;
                    return true;
                }
            }
        }
        if(i2 < s2->length())
        {
            if(s3->at(i3) == s2->at(i2))
            {
                tag1 = check(s1,s2,s3,i1,i2+1,i3+1);
                if(tag1 == true)
                {
                    mark[i1][i2] = 1;
                    return true;
                }
            }
        }
        mark[i1][i2] = 0;
        return false;
    }
};