技術標籤：leetcodeleetcode演算法

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.

Constraints:
n == graph.length
2 <= n <= 15
0 <= graph[i][j] < n
graph[i][j] != i (i.e., there will be no self-loops).
The input graph is guaranteed to be a DAG.

有向無環圖中有n個node，找出0到n-1節點的所有路徑

思路：
從0出發DFS

class Solution {
    int n = 0;
    List<List<Integer>> result;
    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        result = new ArrayList<List<Integer>>();
        n = graph.length;
        boolean[]
 
 visited = new boolean[n];
        List<Integer> tmp = new ArrayList<>();
        
        dfs(graph, tmp, visited, 0);
        return result;
    }
    
    void dfs(int[][] graph, List<Integer> list, boolean[] visited, int node) {
        if(node == n-1) {
            list.add(node);
            result.add(new ArrayList<>(list));
            list.remove(list.size()-1);
            return;
        }
        if(visited[node]) return;
        
        visited[node] = true;
        list.add(node);
        for(int nextNode : graph[node]) {
            dfs(graph, list, visited, nextNode);
        }
        visited[node] = false;
        list.remove(list.size() - 1);
        
    }
}

由於不用遍歷所有節點，而且是沒有環的有向圖，可以不用visited陣列記訪問過的位置

class Solution {
    int n = 0;
    List<List<Integer>> result;
    public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
        result = new ArrayList<List<Integer>>();
        n = graph.length;
        //boolean[] visited = new boolean[n];
        List<Integer> tmp = new ArrayList<>();
        
        dfs(graph, tmp, 0);
        return result;
    }
    
    void dfs(int[][] graph, List<Integer> list, int node) {
        //visited[node] = true;
        list.add(node);
        
        if(node == n-1) {
            result.add(new ArrayList<>(list));
        } else {
           for(int nextNode : graph[node]) {
             dfs(graph, list, nextNode);
           }
        } 
        
        //visited[node] = false;
        list.remove(list.size() - 1);
        
    }
}