本文主要是介绍LeetCode841. Keys and Rooms，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
文章目录
- 一、题目
- 二、题解
一、题目
There are n rooms labeled from 0 to n - 1 and all the rooms are locked except for room 0. Your goal is to visit all the rooms. However, you cannot enter a locked room without having its key.
When you visit a room, you may find a set of distinct keys in it. Each key has a number on it, denoting which room it unlocks, and you can take all of them with you to unlock the other rooms.
Given an array rooms where rooms[i] is the set of keys that you can obtain if you visited room i, return true if you can visit all the rooms, or false otherwise.
Example 1:
Input: rooms = [[1],[2],[3],[]]
Output: true
Explanation:
We visit room 0 and pick up key 1.
We then visit room 1 and pick up key 2.
We then visit room 2 and pick up key 3.
We then visit room 3.
Since we were able to visit every room, we return true.
Example 2:
Input: rooms = [[1,3],[3,0,1],[2],[0]]
Output: false
Explanation: We can not enter room number 2 since the only key that unlocks it is in that room.
Constraints:
n == rooms.length
2 <= n <= 1000
0 <= rooms[i].length <= 1000
1 <= sum(rooms[i].length) <= 3000
0 <= rooms[i][j] < n
All the values of rooms[i] are unique.
二、题解
class Solution {
public:void dfs(vector<vector<int>>& rooms,vector<bool>& visited,int key){if(visited[key]) return;visited[key] = true;vector<int> keys = rooms[key];for(int i = 0;i < keys.size();i++){dfs(rooms,visited,keys[i]);}}bool canVisitAllRooms(vector<vector<int>>& rooms) {vector<bool> visited(rooms.size(),false);dfs(rooms,visited,0);for(int i = 0;i < visited.size();i++){if(visited[i] == false) return false;}return true;}
};
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