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.

題目看了半天才看懂，大意就是：

給出了一個有向無環圖，求從起點到終點的所有路徑。圖的表示方法是，共有n個節點，其數字分別為0…n-1，給出的圖graph的每個位置對應的是第i個節點能到達的下一個節點的序號位置。比如題中graph[0] = [1,2]表示圖的起點0指向了1,2兩個節點。

思路是利用DFS演算法

class Solution {
public:
   bool getAllPath(vector<vector<int>>& graph,int node,vector<vector<int>> &ret,vector<int> &path){
        /*the last node*/
        if(node == graph.size()-1){
            ret.push_back(path);
            return true;
        }
        
        /*push the first node*/
        for(int i = 0;i < graph[node].size();++i){
            path.push_back(graph[node][i]);
            getAllPath(graph,graph[node][i],ret,path);
            path.pop_back();
        }
    
        return true;
    }
 
    vector<vector<int>> allPathsSourceTarget(vector<vector<int>>& graph) {
        stack<int> s;
        int n = graph.size();
        vector<vector<int>> ret;
        vector<int> path;
        
        if(graph.size() <= 0){
            return ret;
        }
        
        path.push_back(0);
        getAllPath(graph,0,ret,path);
        return ret;
    }
};