Given a directed, acyclic graph ofNnodes. Find all possible paths from node0to nodeN-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.

Backtracking，由于可能存在环路，所以要使用一个boolean数组记录各个点是否已经遍历过。

class Solution {
    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        List<List<Integer>> res = new ArrayList<>();
        if(graph == null || graph.length == 0)
            return res;

        List<Integer> line = new ArrayList<Integer>();
        boolean[] visited = new boolean[graph.length];
        line.add(0);
        visited[0] = true;
        helper(res, line, 0, visited, graph, graph.length - 1);

        return res;
    }

    public void helper(List<List<Integer>> res, List<Integer> line, int start, boolean[] visited, int[][] graph, int target) {
        if(start == target) {
            res.add(new ArrayList<Integer>(line));
            return;
        }

        int[] next = graph[start];
        for(int n: next) {
            if(visited[n])
                continue;

            line.add(n);
            visited[n] = true;
            helper(res, line, n, visited, graph, target);
            visited[n] = false;
            line.remove(line.size() - 1);
        }
    }
}