797. All Paths from Source to Target

Given a directed acyclic graph (DAG) of n nodes labeled from 0 to n - 1, find all possible paths from node 0 to node n - 1 and return them in any order.

The graph is given as follows: graph[i] is a list of all nodes you can visit from node i (i.e., there is a directed edge from node i to node graph[i][j]).

Example 1:

Input: graph = [[1,2],[3],[3],[]]
Output: [[0,1,3],[0,2,3]]
Explanation: There are two paths: 0 -> 1 -> 3 and 0 -> 2 -> 3.

Example 2:

Input: graph = [[4,3,1],[3,2,4],[3],[4],[]]
Output: [[0,4],[0,3,4],[0,1,3,4],[0,1,2,3,4],[0,1,4]]

Constraints:

• n == graph.length

• 2 <= n <= 15

• 0 <= graph[i][j] < n

• graph[i][j] != i (i.e., there will be no self-loops).

• All the elements of graph[i] are unique.

• The input graph is guaranteed to be a DAG.

My Solutions:

用dfs的方法，从起点0开始dfs遍历ta可以走到的每一条路。如果source==target，说明发现一条路径，把路径加入结果。

class Solution {

    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        List<List<Integer>> res = new ArrayList<>();
        List<Integer> currPath = new ArrayList<>();
        currPath.add(0); // 先把0放进去，因为0是source
        int n = graph.length - 1; // 最后一个数字是target
        dfs(graph, currPath, 0, n, res);
        return res;
    }

    private void dfs(int[][] graph, List<Integer> currPath, int source, int target, List<List<Integer>> res) {
        if (source == target)  {
            res.add(new ArrayList<>(currPath)); // 注意这里！把路径加入结果
            return;
        }
        for (int i : graph[source]) { // iterate all the destination nodes for current source node
             currPath.add(i);
             dfs(graph, currPath, i, target, res);
             currPath.remove(currPath.size() - 1);
        }
    }

}