leetcode 797. All Paths From Source to Target(python3)


leetcode 797. All Paths From Source to Target(python3)


原题地址:https://leetcode.com/problems/all-paths-from-source-to-target/

题目

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.

分析

本题使用的思想是DFS深度优先搜索,从0索引开始,根据列表里面的值跳到相应索引,直到遇到终点值返回,并根据此次搜索是否到达该列表末尾选择pop出末尾值。

代码块

class Solution:
    def allPathsSourceTarget(self, graph):
        """ :type graph: List[List[int]] :rtype: List[List[int]] """
        for i in graph:
            i.sort()
        fianl = []  #存储最终返回的列表
        a = [0]  #每次搜索均从0开始
        def dfs(n,a):
            for i in graph[n]:
                a.append(i)
                if i == len(graph)-1:
                    fianl.append(a[:])
                    a.pop()
                    print(a)
                    if i == graph[n][-1]:   #此次搜索到达该列表末尾,弹出末尾值
                        a.pop()
                        print(a)
                    return
                else:
                    dfs(i, a)
                    if i == graph[n][-1]:
                        a.pop()

        dfs(0, a)
        return fianl

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