Given a directed, acyclic graph of N nodes. Find all possible paths from node 0 to node N-1, and return them in any order.

The graph is given as follows: the nodes are 0, 1, …, graph.length - 1. graph[i] is a list of all nodes j for which the edge (i, j) exists.

Example:

Input: [[1,2], [3], [3], []] 
Output: [[0,1,3],[0,2,3]] 
Explanation: The graph looks like this:
0--->1
|    |
v    v
2--->3
There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Note:

The number of nodes in the graph will be in the range [2, 15].
You can print different paths in any order, but you should keep the order of nodes inside one path.

solution

time complexity : O(n!)
space complexity : O(n)

class Solution {
public:
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        vector<int> curRes = {0};
        dfs(curRes, graph[0], graph);
        return res;
    }
private:
    vector<vector<int>> res;
    void dfs(vector<int> &curRes, vector<int> nexts, vector<vector<int>>& graph) {
        if (0 == nexts.size()) {
            res.push_back(curRes);
            return;
        }
        
        for (int num: nexts) {
            curRes.push_back(num);
            dfs(curRes, graph[num], graph);
            curRes.pop_back();
        }
    }
};

題目的輸入意思是：第一個node 0指向node 1、2，…以此類推。
列舉從頭到尾的路線，利用dfs以及backtracking，即可一一得出答案。