# 797. All Paths From Source to Target

## 1. Question

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]).

## 2. Examples

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]]


Example 3:

Input: graph = [[1],[]]
Output: [[0,1]]


Example 4:

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


Example 5:

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


## 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).
• All the elements of graph[i] are unique.
• The input graph is guaranteed to be a DAG.

## 5. Solutions

dfs

class Solution {
List<List<Integer>> lists = new ArrayList<>();
public List<List<Integer>> allPathsSourceTarget(int[][] graph) {
dfs(graph, 0, new ArrayList<>());
return lists;
}

public void dfs(int[][] graph, int index, List<Integer> list) {
if (index == graph.length - 1) {
return ;
}
for (int i : graph[index]) {
dfs(graph, i, list);
list.remove(list.size() - 1);   // 不符合条件的要删掉
}

}
}