97. Interleaving String (String; DP)

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        if(s1.empty())
        {
            if(s2 == s3) return true;
            else return false;
        }
        else if(s2.empty())
        {
            if(s1==s3) return true;
            else return false;
        }
        
        int l1 = s1.length();
        int l2 = s2.length();
        int l3 = s3.length();
        if(l3!=l1+l2) return false;
        
        bool** hit = new bool* [l1+1]; //状态hit[i][j]表示s3[i+j-1]由s1的前i个字符和s2的前j个字符组成
        for(int i = 0; i<=l1; i++)
        {
            hit[i] = new bool [l2+1];
        }
        
        //initialize state
        hit[0][0] = true;
        for ( int i = 1; i<=l1; i++)
        {
            if(s1[i-1] == s3[i-1])
            {
                hit[i][0] = true;
            }
            else
            {
                for(int j = i; j <= l1; j++){
                    hit[j][0] = false;
                }
                break;
            }
        }
        for ( int i = 1; i<=l2; i++)
        {
            if(s2[i-1] == s3[i-1])
            {
                hit[0][i] = true;
            }
            else
            {
                for(int j = i; j <= l2; j++){
                    hit[0][j] = false;
                }
                break;
            }
        }
        
        //状态转移
        for ( int i = 1; i<=l1; i++)
        {
            for( int j = 1; j<=l2; j++)
            {
                if(s1[i-1]==s3[i+j-1])
                {
                    hit[i][j] = hit[i-1][j];
                }
                if(s2[j-1]==s3[i+j-1])
                {
                    hit[i][j] = hit[i][j-1] || hit[i][j];
                }
            }
        }
        return hit[l1][l2];
    }
};