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[LeetCode] 1845. Seat Reservation Manager

开发技术 开发技术 2周前 (05-01) 5次浏览

Design a system that manages the reservation state of n seats that are numbered from 1 to n.

Implement the SeatManager class:

  • SeatManager(int n) Initializes a SeatManager object that will manage n seats numbered from 1 to n. All seats are initially available.
  • int reserve() Fetches the smallest-numbered unreserved seat, reserves it, and returns its number.
  • void unreserve(int seatNumber) Unreserves the seat with the given seatNumber.

Example 1:

Input
["SeatManager", "reserve", "reserve", "unreserve", "reserve", "reserve", "reserve", "reserve", "unreserve"]
[[5], [], [], [2], [], [], [], [], [5]]
Output
[null, 1, 2, null, 2, 3, 4, 5, null]

Explanation
SeatManager seatManager = new SeatManager(5); // Initializes a SeatManager with 5 seats.
seatManager.reserve();    // All seats are available, so return the lowest numbered seat, which is 1.
seatManager.reserve();    // The available seats are [2,3,4,5], so return the lowest of them, which is 2.
seatManager.unreserve(2); // Unreserve seat 2, so now the available seats are [2,3,4,5].
seatManager.reserve();    // The available seats are [2,3,4,5], so return the lowest of them, which is 2.
seatManager.reserve();    // The available seats are [3,4,5], so return the lowest of them, which is 3.
seatManager.reserve();    // The available seats are [4,5], so return the lowest of them, which is 4.
seatManager.reserve();    // The only available seat is seat 5, so return 5.
seatManager.unreserve(5); // Unreserve seat 5, so now the available seats are [5].

Constraints:

  • 1 <= n <= 105
  • 1 <= seatNumber <= n
  • For each call to reserve, it is guaranteed that there will be at least one unreserved seat.
  • For each call to unreserve, it is guaranteed that seatNumber will be reserved.
  • At most 105 calls in total will be made to reserve and unreserve.

座位预约管理系统。

请你设计一个管理 n 个座位预约的系统,座位编号从 1 到 n 。

请你实现 SeatManager 类:

SeatManager(int n) 初始化一个 SeatManager 对象,它管理从 1 到 n 编号的 n 个座位。所有座位初始都是可预约的。
int reserve() 返回可以预约座位的 最小编号 ,此座位变为不可预约。
void unreserve(int seatNumber) 将给定编号 seatNumber 对应的座位变成可以预约。

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/seat-reservation-manager
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这是一道设计题,我这里提供一个最小堆的做法。注意题目标红的条件,unreserve的座位号一定是之前已经被reserve过的。没有这个条件,这个题是没法做的。我直接给代码了。

时间O(nlogn)

空间O(n)

Java实现

 1 class SeatManager {
 2     PriorityQueue<Integer> minHeap;
 3 
 4     public SeatManager(int n) {
 5         minHeap = new PriorityQueue<>();
 6         for (int i = 1; i <= n; i++) {
 7             minHeap.offer(i);
 8         }
 9     }
10 
11     public int reserve() {
12         if (!minHeap.isEmpty()) {
13             return minHeap.poll();
14         }
15         return -1;
16     }
17 
18     public void unreserve(int seatNumber) {
19         minHeap.offer(seatNumber);
20     }
21 }
22 
23 /**
24  * Your SeatManager object will be instantiated and called as such:
25  * SeatManager obj = new SeatManager(n);
26  * int param_1 = obj.reserve();
27  * obj.unreserve(seatNumber);
28  */

 

LeetCode 题目总结


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