797. All Paths From Source to Target

題目

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

一開始也沒看懂題意，這個二維數組是如何描述這個有向圖。其實是這樣的，二維數組的長度就是節點的個數，比如示例中的長度爲4，所以節點分爲是0, 1, 2, 3。
然而，下一維度描述了，當前第i個節點指向哪個節點。
比如，graph[0]是[1, 2]，他的意思是，0號節點指向1,2

class P797_All_Paths_From_Source_to_Target {

    // 把結果集弄成靜態變量，省的傳來傳去
    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) {
        
        // 將當前節點加入路徑
        path.add(curr);

        // 查一下當前節點的鄰接表，如果長度爲0，所以到頭了，把這條路徑加入結果集
        if (graph[curr].length == 0) {
            result.add(path);
        } else {
            // 如果沒有到頭，那麼查鄰接表，每一個都取出來進行遞歸。
            for (int n : graph[curr]) {
                // 這裏要注意克隆了一下路徑數組，不然的話，都加到一個裏面去了……
                helper(n, graph, clone(path));
            }
        }

    }

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

}