797 All Paths From Source to Target

797. All Paths From Source to Target

1. Question

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.

2. Implementation

(1) Backtracking

思路: 题目要求的是从Node 0 到 Node n - 1的所有路径，所以base case里，当node == n - 1时，我们就会找到一个可行解，通过backtracking找到所有路径

class Solution {
    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        List<List<Integer>> res = new ArrayList<>();

        if (graph == null || graph.length == 0) {
            return res;
        }

        int m = graph.length, n = graph[0].length;

        List<Integer> path = new ArrayList<>();
        path.add(0);
        getPaths(0, graph, path, res);
        return res;
    }

    public void getPaths(int curNode, int[][] graph, List<Integer> path, List<List<Integer>> res) {
        if (curNode == graph.length - 1) {
            res.add(new ArrayList<>(path));
            return;
        }

        for (int nextNode : graph[curNode]) {
            path.add(nextNode);
            getPaths(nextNode, graph, path, res);
            path.remove(path.size() - 1);
        }
    }
}

3. Time & Space Complexity

Backtracking: 时间复杂度O(N * 2^N ), 总共有2^N个可能的路径, 而每个路径path.remove()需要O(N)的时间, 空间复杂度O(N * 2^N)，总共有2 ^ N的路径，每个路径的递归深度最多为N