[leetcode] 797. All Paths From Source to Target

Description

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).
• The input graph is guaranteed to be a DAG.

分析

题目的意思是：给定一个有向图，找出所有从0到n-1的路径。思路也很直接，就是深度优先搜索。

• 终止条件就是node节点到达n-1位置，直接返回当前节点就行了
• 然后对于当前的相邻节点进行遍历，对于遍历到的节点进行递归，这样下去就可以找到所有到达n-1节点的路径了。

代码

class Solution:
    def solve(self,graph,node,n):
        if(node==n-1):
            return [[n-1]]
        res=[]
        for n1 in graph[node]:
            for path in self.solve(graph,n1,n):
                res.append([node]+path)
        return res
    def allPathsSourceTarget(self, graph: List[List[int]]) -> List[List[int]]:
        n=len(graph)
        res=self.solve(graph,0,n)
        return res

参考文献

所有可能的路径