動機

就dfs，同時因為是dag所以不用處理cycle

Problem

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]]

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]]

 

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.

Sol

class Solution:
    def allPathsSourceTarget(self, gh: List[List[int]]) -> List[List[int]]:
        def dfs(n, path):
            if n == len(gh)-1:
                return [path+[n]]
            else:
                ret = []
                for x in gh[n]:
                    ret += dfs(x, path+[n])
                return ret
        return dfs(0,[])