敢教日月换新天。为有牺牲多壮志,

[Swift]LeetCode841. 钥匙和房间 | Keys and Rooms

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号:山青咏芝(shanqingyongzhi)
➤博客园地址:山青咏芝(https://www.cnblogs.com/strengthen/
➤GitHub地址:https://github.com/strengthen/LeetCode
➤原文地址: https://www.cnblogs.com/strengthen/p/10579789.html 
➤如果链接不是山青咏芝的博客园地址,则可能是爬取作者的文章。
➤原文已修改更新!强烈建议点击原文地址阅读!支持作者!支持原创!
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

热烈欢迎,请直接点击!!!

进入博主App Store主页,下载使用各个作品!!!

注:博主将坚持每月上线一个新app!!!

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 [0, 1, ..., N-1] where N = rooms.length.  A key rooms[i][j] = v opens the room with number v.

Initially, all the rooms start locked (except for room 0). 

You can walk back and forth between rooms freely.

Return true if and only if you can enter every room.

Example 1:

Input: [[1],[2],[3],[]]
Output: true
Explanation:  
We start in room 0, and pick up key 1.
We then go to room 1, and pick up key 2.
We then go to room 2, and pick up key 3.
We then go to room 3.  Since we were able to go to every room, we return true.

Example 2:

Input: [[1,3],[3,0,1],[2],[0]]
Output: false
Explanation: We can't enter the room with number 2.

Note:

  1. 1 <= rooms.length <= 1000
  2. 0 <= rooms[i].length <= 1000
  3. The number of keys in all rooms combined is at most 3000.

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。

如果能进入每个房间返回 true,否则返回 false

示例 1:

输入: [[1],[2],[3],[]]
输出: true
解释:  
我们从 0 号房间开始,拿到钥匙 1。
之后我们去 1 号房间,拿到钥匙 2。
然后我们去 2 号房间,拿到钥匙 3。
最后我们去了 3 号房间。
由于我们能够进入每个房间,我们返回 true。

示例 2:

输入:[[1,3],[3,0,1],[2],[0]]
输出:false
解释:我们不能进入 2 号房间。

提示:

  1. 1 <= rooms.length <= 1000
  2. 0 <= rooms[i].length <= 1000
  3. 所有房间中的钥匙数量总计不超过 3000

40ms

复制代码
 1 class Solution {
 2     func canVisitAllRooms(_ rooms: [[Int]]) -> Bool {
 3         var keys: [Int] = Array(repeating: -1, count: rooms.count)
 4         var result:[Int] = [0]
 5         keys[0] = 0
 6         var i: Int = 0
 7         while i < result.count {
 8             for each in rooms[result[i]] {
 9                 if keys[each] == -1 {
10                     result.append(each)
11                     keys[each] = 0
12                 }
13             }
14             if result.count == rooms.count {
15                 return true
16             }
17             i += 1
18         }
19         return false
20     }
21 }
复制代码

Runtime: 44 ms
Memory Usage: 19.4 MB
复制代码
 1 class Solution {
 2     func canVisitAllRooms(_ rooms: [[Int]]) -> Bool {
 3         var dfs:[Int] = [Int]()
 4         dfs.append(0)
 5         var seen:Set<Int> = Set<Int>()
 6         seen.insert(0)
 7         while(!dfs.isEmpty)
 8         {
 9             var i:Int = dfs.removeLast()
10             for j in rooms[i]
11             {
12                 if !seen.contains(j)
13                 {
14                     dfs.append(j)
15                     seen.insert(j)
16                     if rooms.count == seen.count {return true}
17                 }
18             }
19         }
20         return rooms.count == seen.count
21     }
22 }
复制代码

44ms

复制代码
 1 class Solution {
 2     func canVisitAllRooms(_ rooms: [[Int]]) -> Bool {
 3         let count = rooms.count
 4         var seen = [Bool](repeating: false, count: count)
 5         seen[0] = true
 6         
 7         var keysInRoom = [Int]()
 8         keysInRoom.append(0)
 9         
10         while keysInRoom.count != 0 {
11             let room = keysInRoom.last!
12             keysInRoom.removeLast()
13             
14             let keys = rooms[room]
15             for key in keys {
16                 if !seen[key] {
17                     seen[key] = true
18                     keysInRoom.append(key)
19                 }
20             }
21         }
22         for s in seen {
23             if !s {
24                 return false
25             }
26         }
27         return true
28     }
29 }
复制代码

48ms

复制代码
 1 class Solution {
 2     func canVisitAllRooms(_ rooms: [[Int]]) -> Bool {
 3 
 4         var numberOfRooms = rooms.count
 5         var visited: [Int: Int] = [:]
 6         var stack: [Int] = []
 7         stack.append(0)
 8         visited[0] = 1
 9         
10         while(!stack.isEmpty) {
11             
12             var room = stack.removeLast()
13             var keys = rooms[room]
14             
15             for key in keys {
16                 if(visited[key] == nil) {
17                     visited[key] = 1
18                     stack.append(key)
19                 }
20             }            
21         }
22         return numberOfRooms == visited.keys.count
23     }
24 }
复制代码

52ms

复制代码
 1 class Solution {
 2     func canVisitAllRooms(_ rooms: [[Int]]) -> Bool {
 3         var key = Set(rooms[0])
 4         var open = [0:0]
 5         var nextOpen : Int? = 0
 6         while nextOpen != nil {
 7             nextOpen = nil
 8             for data in key {
 9                 if open[data] == nil {
10                 nextOpen = data
11                open[ data] = data
12                 for k in rooms[data]{
13                     key.insert(k)
14                  }
15                 key.remove(data)
16                 }
17             }
18         }
19         return open.count == rooms.count
20     }
21 }
复制代码

64ms

复制代码
 1 class Solution {
 2     func canVisitAllRooms(_ rooms: [[Int]]) -> Bool {
 3         var unvisited : Set<Int> = []
 4         var canVisit : [Int] = []
 5         
 6         for i in 0 ..< rooms.count {
 7             unvisited.insert(i)
 8         }
 9         
10         canVisit = rooms[0]
11         unvisited.remove(0)
12         
13         while !canVisit.isEmpty {
14             let room = canVisit.removeFirst()
15             if unvisited.contains(room) {
16                 canVisit.append(contentsOf: rooms[room])
17                 unvisited.remove(room)
18             }
19         }
20         
21         if unvisited.isEmpty {
22             return true
23         }
24         
25         return false
26     }
27 }
复制代码

76ms

复制代码
 1 class Solution {
 2     func canVisitAllRooms(_ rooms: [[Int]]) -> Bool {
 3         if (rooms.count <= 1) { return true }
 4         
 5         var visitedRooms: [Bool] = Array.init(repeating: false, count: rooms.count)
 6         visitedRooms[0] = true
 7         var queue: [Int] = []
 8         addKeysToQueue(&queue, 0, &visitedRooms, rooms)
 9         
10         while (!queue.isEmpty) {
11             addKeysToQueue(&queue, queue.removeLast(), &visitedRooms, rooms)
12         }
13         
14         for i in 0..<visitedRooms.count {
15             if (!visitedRooms[i]) {
16                 return false
17             }
18         }
19         return true
20     }
21     
22     func addKeysToQueue(_ queue: inout [Int], _ key: Int, _ visitedRooms: inout [Bool], _ rooms: [[Int]]) {
23         for i in 0..<rooms[key].count {
24             var nextRoom = rooms[key][i]
25             if (!visitedRooms[nextRoom]) {
26                 queue.append(nextRoom)
27                 visitedRooms[nextRoom] = true
28             }
29         }
30     }
31 }
复制代码

 

 

posted @   为敢技术  阅读(302)  评论(0编辑  收藏  举报
编辑推荐:
· 记一次.NET内存居高不下排查解决与启示
· 探究高空视频全景AR技术的实现原理
· 理解Rust引用及其生命周期标识(上)
· 浏览器原生「磁吸」效果!Anchor Positioning 锚点定位神器解析
· 没有源码,如何修改代码逻辑?
阅读排行:
· 全程不用写代码,我用AI程序员写了一个飞机大战
· MongoDB 8.0这个新功能碉堡了,比商业数据库还牛
· DeepSeek 开源周回顾「GitHub 热点速览」
· 记一次.NET内存居高不下排查解决与启示
· 白话解读 Dapr 1.15:你的「微服务管家」又秀新绝活了
点击右上角即可分享
微信分享提示
哥伦布
09:09发布
哥伦布
09:09发布
3°
多云
东南风
3级
空气质量
相对湿度
47%
今天
中雨
3°/15°
周三
中雨
3°/13°
周四
小雪
-1°/6°