797. All Paths From Source to Target

题目

https://leetcode.com/problems/all-paths-from-source-to-target/

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.

解答

一开始也没看懂题意，这个二维数组是如何描述这个有向图。其实是这样的，二维数组的长度就是节点的个数，比如示例中的长度为4，所以节点分为是0, 1, 2, 3。
然而，下一维度描述了，当前第i个节点指向哪个节点。
比如，graph[0]是[1, 2]，他的意思是，0号节点指向1,2

class P797_All_Paths_From_Source_to_Target {

    // 把结果集弄成静态变量，省的传来传去
    static List<List<Integer>> result = new ArrayList<>();

    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        List<Integer> path = new ArrayList<>();
        result.clear();
        helper(0, graph, path);
        return result;
    }

    public static void helper(int curr, int[][] graph, List<Integer> path) {
        
        // 将当前节点加入路径
        path.add(curr);

        // 查一下当前节点的邻接表，如果长度为0，所以到头了，把这条路径加入结果集
        if (graph[curr].length == 0) {
            result.add(path);
        } else {
            // 如果没有到头，那么查邻接表，每一个都取出来进行递归。
            for (int n : graph[curr]) {
                // 这里要注意克隆了一下路径数组，不然的话，都加到一个里面去了……
                helper(n, graph, clone(path));
            }
        }

    }

    public static List<Integer> clone(List<Integer> arr) {
        return new ArrayList<>(arr);
    }

}