There are N rooms and you start in room 0.  Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room. Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an integer in […

原题链接在这里:https://leetcode.com/problems/keys-and-rooms/ 题目: There are N rooms and you start in room 0.  Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room. Formally, each room i has a list of…

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 DFS BFS 日期 题目地址:https://leetcode.com/problems/keys-and-rooms/description/ 题目描述 There are N rooms and you start in room 0. Each room has a distinct number in 0, 1, 2, -, N-1, an…

题目链接:https://leetcode.com/problems/keys-and-rooms/description/ 简单DFS time:9ms 1 class Solution { 2 public: 3 void DFS(int root,vector<int>& visited,vector<vector<int>>& rooms){ 4 visited[root] = 1; 5 for(auto x : rooms[root]){ 6…

题目: ​ 有 N 个房间,开始时你位于 0 号房间.每个房间有不同的号码:0,1,2,...,N-1,并且房间里可能有一些钥匙能使你进入下一个房间. ​ 在形式上,对于每个房间 i 都有一个钥匙列表 rooms[i],每个钥匙 rooms[i][j] 由 [0,1,...,N-1] 中的一个整数表示,其中 N = rooms.length. 钥匙 rooms[i][j] = v 可以打开编号为 v 的房间. 最初,除 0 号房间外的其余所有房间都被锁住. 你可以自由地在房间之间来回走动. 如果…

There are N rooms and you start in room 0.  Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room. Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an integer in […

There are N rooms and you start in room 0.  Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room. Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an integer in […

""" There are N rooms and you start in room 0. Each room has a distinct number in 0, 1, 2, ..., N-1, and each room may have some keys to access the next room. Formally, each room i has a list of keys rooms[i], and each key rooms[i][j] is an…

1.题目描述 2.问题分析 使用深度优先遍历 3.代码 bool canVisitAllRooms(vector<vector<int>>& rooms) { int nums = rooms.size(); ) return true; vector<); R[] = ; stack<int> s; s.push(); while (!s.empty()) { int room = s.top(); s.pop(); for(auto n : rooms…

DFS基础 深度优先搜索(Depth First Search)是一种搜索思路,相比广度优先搜索(BFS),DFS对每一个分枝路径深入到不能再深入为止,其应用于树/图的遍历.嵌套关系处理.回溯等,可以用递归.堆栈(stack)实现DFS过程. 关于广度优先搜索(BFS)详见:算法与数据结构基础 - 广度优先搜索(BFS) 关于递归(Recursion)详见:算法与数据结构基础 - 递归(Recursion) 树的遍历 DFS常用于二叉树的遍历,关于二叉树详见: 算法与数据结构基础 - 二叉查找树…