1. 题目

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.

2. 思路

递归的思路比较好理解，第一版直接递归，超时了。
加入了结果的cache，在已经计算过的结果里，降低递归的次数，AC。
cache用一个flagi的数组来标识，初始为2，表示还没计算过。如果计算过，则赋值为0或1，代表不匹配和匹配。

3. 代码

耗时：6ms

class Solution {
    int flag[512][512];
public:
    bool isInterleave(string s1, string s2, string s3) {
        int l1 = s1.length();
        int l2 = s2.length();
        int l3 = s3.length();
        if (l1+l2!=l3) { return false; }
        for (int i = 0; i < l1+1; i++) {
            for (int j = 0; j < l2+1; j++) {
                flag[i][j] = 2;
            }
        }
        return isInterleave(s1, 0, s2, 0, s3, 0);
    }
    
    bool isInterleave(string& si, int i, string& sj, int j, string& sk, int k) {
        if (i == si.length()) { flag[i][j] = sj.substr(j) == sk.substr(k) ? 1 : 0; return flag[i][j] == 1; }
        if (j == sj.length()) { flag[i][j] = si.substr(i) == sk.substr(k) ? 1 : 0; return flag[i][j] == 1; }
        int r_i = i;
        int r_j = j;
        while (si[i] != sj[j] && i < si.length() && j < sj.length()) {
            if (si[i] == sk[k]) {
                i++; k++;
            } else if (sj[j] == sk[k]) {
                j++; k++;
            } else {
                flag[r_i][r_j] = 0;
                flag[i][j] = 0;
                return false;
            }
        }
        if (i == si.length() || j == sj.length()) {
            return isInterleave(si, i, sj, j, sk, k);
        }
        if (si[i] == sk[k]) {
            if (flag[i+1][j] == 2) {
                flag[i+1][j] = isInterleave(si, i+1, sj, j, sk, k+1) ? 1 : 0;
            }
            if (flag[i+1][j] == 1) { return true; }
            if (flag[i][j+1] == 2) {
                flag[i][j+1] = isInterleave(si, i, sj, j+1, sk,k+1) ? 1 : 0;
            }
            if (flag[i][j+1] == 1) { return true; }
        }
        flag[i][j] = 0;
        return false;
    }
};