LeetCode-Interleaving String

很显然的DP，基本算是一次过了，

比较简单！

class Solution {
public:
    bool isInterleave(string s1, string s2, string s3) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function    
        int n1 = s1.size();
        int n2 = s2.size();
        int n3 = s3.size();
        vector<vector<int> > sat;
        vector<int> temp(n2, -1);
        for (int i = 0; i < n1; ++i) {
            sat.push_back(temp);
        }
        return inter(sat, s1, 0, s2, 0, s3, 0);
    }
    bool inter(vector<vector<int> > &sat, string &s1, int b1, string &s2, int b2, string &s3, int b3) {
        if (b1 == s1.size() && b2 == s2.size() && b3 == s3.size()) {
            return true;
        }
        if ((s1.size() + s2.size()) != s3.size()) {
            return false;
        }
        if (b1 == s1.size()) {
            string ts2 = s2.substr(b2, s2.size() - b2);
            string ts3 = s3.substr(b3, s3.size() - b3);
            if (ts2 == ts3) {
                return true;
            }
            return false;
        }
        if (b2 == s2.size()) {
            string ts1 = s1.substr(b1, s1.size() - b1);
            string ts3 = s3.substr(b3, s3.size() - b3);
            if (ts1 == ts3) {
                return true;
            }
            return false;
        }
        if (sat[b1][b2] != -1) {
            return sat[b1][b2];
        }
        if (s1[b1] != s2[b2]) {
            if (s3[b3] == s1[b1]) {
                bool res = inter(sat, s1, b1 + 1, s2, b2, s3, b3 + 1);
                sat[b1][b2] = res;
                return res;
            }
            if (s3[b3] == s2[b2]) {
                bool res = inter(sat, s1, b1, s2, b2 + 1, s3, b3 + 1);
                sat[b1][b2] = res;
                return res;
            }
            sat[b1][b2] = false;
            return false;
        }
        else {
            if (s3[b3] == s1[b1]) {
                bool res = inter(sat, s1, b1 + 1, s2, b2, s3, b3 + 1) || inter(sat, s1, b1, s2, b2 + 1, s3, b3 + 1);
                sat[b1][b2] = res;
                return res;
            }
            sat[b1][b2] = false;
            return false;
        }
    }
};