题目链接: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.

看完题目描述,直觉就是DFS搜索解空间树。因为不是二维表格所以不好用DP,同时是无环图所以感觉比较像DFS。代码如下:

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)

感觉是一道很标准的DFS,没有什么难点。

LeetCode 797. All Paths From Source to Target的更多相关文章

  1. 【LeetCode】797. All Paths From Source to Target 解题报告(Python & C++)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 回溯法 日期 题目地址:https://leetco ...

  2. 【leetcode】797. All Paths From Source to Target

    Given a directed acyclic graph (DAG) of n nodes labeled from 0 to n - 1, find all possible paths fro ...

  3. 【leetcode】All Paths From Source to Target

    题目如下: Given a directed, acyclic graph of N nodes. Find all possible paths from node 0 to node N-1, a ...

  4. LeetCode 1059. All Paths from Source Lead to Destination

    原题链接在这里:https://leetcode.com/problems/all-paths-from-source-lead-to-destination/ 题目: Given the edges ...

  5. [LeetCode] All Paths From Source to Target 从起点到目标点到所有路径

    Given a directed, acyclic graph of N nodes.  Find all possible paths from node 0 to node N-1, and re ...

  6. 75th LeetCode Weekly Contest All Paths From Source to Target

    Given a directed, acyclic graph of N nodes.  Find all possible paths from node 0 to node N-1, and re ...

  7. [Swift]LeetCode797. 所有可能的路径 | All Paths From Source to Target

    Given a directed, acyclic graph of N nodes.  Find all possible paths from node 0 to node N-1, and re ...

  8. LeetCode 63. Unique Paths II不同路径 II (C++/Java)

    题目: A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). ...

  9. [LeetCode] 62. Unique Paths 唯一路径

    A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). The ...

随机推荐

  1. ssm登录与退出

    ssm整合比较好的实例 http://how2j.cn/k/ssm/ssm-tutorial/1137.html?tid=77#nowhere ssm登录后台用户检测(此实例注销有问题):http:/ ...

  2. 忽略SIGPIPE信号

    #include <stdlib.h> #include <sys/signal.h> void SetupSignal() { struct sigaction sa; // ...

  3. 小伙 zwfw-new.hunan.gov.cn.iname.damddos.com [222.240.80.52]

    由于这个应用出问题非常影响用户体验:于是立马让运维保留现场 dump 线程和内存同时重启应用,还好重启之后恢复正常.于是开始着手排查问题.

  4. vim 加密(crypt)文本文档

    算法 vim7.3版本支持两种加密方式——PKzip算法(已知有缺陷的).Blowfish算法(从7.3版本开始支持).Blowfish2算法(从7.4.399版本开始支持)而vim -x 默认采用P ...

  5. java数组的定义

    class ArrayDome { public static void main(String[] args) { //元素类型[] 数组名 创建一个 元素类型[元素个数或数组长度] /* 需要一个 ...

  6. Linux下的tar压缩解压缩命令详解(转)

    tar -c: 建立压缩档案-x:解压-t:查看内容-r:向压缩归档文件末尾追加文件-u:更新原压缩包中的文件 这五个是独立的命令,压缩解压都要用到其中一个,可以和别的命令连用但只能用其中一个.下面的 ...

  7. 腾讯广告联盟 Android SDK(广点通)demo的使用方式

    1. 下载示例文件. 2. 解压之后的目录: 3. 使用android studio,选择import project,导入如图所示文件夹: 4. 重点来了,由于官方demo的上传时间很久远(大概是上 ...

  8. myEclipse出现cannot paste the clipboard contents into the selected elements报错

    导入jar包报错,cannot paste the clipboard contents into the selected elements,查阅资料让重新打开工程,但依然报错. 最后在本地路径复制 ...

  9. Linux 系统的用户和组

    目录 1. 用户及组相关文件 2. 用户相关查询 2.1 直接通过cat文件查看用户及组文件内容 2.2 使用下面查询命令查看 3. 使用操作命令修改用户及组相关文件 3.1 专有编辑命令(仅限高级用 ...

  10. selenium_Python3_邮箱登录:动态元素定位

    这里的关键是动态frame定位: 其他元素定位不用多说,常规操作. 不过需要注意加上这个: from selenium.webdriver.remote.webelement import WebEl ...