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Leetcode 797. All Paths From Source to Target

維表 you length HERE des sso 一道 else rect

題目鏈接:https://leetcode.com/problems/all-paths-from-source-to-target/description/

Given a directed, acyclic graph of N nodes. Find all possible paths from node 0 to node N-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.

Note:

  • The number of nodes in the graph will be in the range [2, 15].
  • You can print different paths in any order, but you should keep the order of nodes inside one path.

看完題目描述,直覺就是DFS搜索解空間樹。因為不是二維表格所以不好用DP,同時是無環圖所以感覺比較像DFS。代碼如下:

class Solution(object):
    def allPathsSourceTarget(self, graph):
        """
        :type graph: List[List[int]]
        :rtype: List[List[int]]
        """
        res = []
        target = len(graph) - 1
        self.dfs([0], res, graph[0], graph, target)
        
return res def dfs(self, curr_sol, res, curr_node, graph, target): if not curr_node: return for nxt in curr_node: if nxt == target: res.append(curr_sol + [nxt]) else: self.dfs(curr_sol+[nxt], res, graph[nxt], graph, target)

感覺是一道很標準的DFS,沒有什麽難點。

Leetcode 797. All Paths From Source to Target