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 aSeatManager
object that will managen
seats numbered from1
ton
. 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 givenseatNumber
.
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 thatseatNumber
will be reserved. - At most
105
calls in total will be made toreserve
andunreserve
.
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 */