LeetCode（Queue)933. Number of Recent Calls

男娘i 2023-09-25 19:33 34阅读 0赞

# 1.问题 #

You have a RecentCounter class which counts the number of recent requests within a certain time frame.

Implement the RecentCounter class:

RecentCounter() Initializes the counter with zero recent requests.  
int ping(int t) Adds a new request at time t, where t represents some time in milliseconds, and returns the number of requests that has happened in the past 3000 milliseconds (including the new request). Specifically, return the number of requests that have happened in the inclusive range \[t - 3000, t\].  
It is guaranteed that every call to ping uses a strictly larger value of t than the previous call.

Example 1:

> Input  
> \[“RecentCounter”, “ping”, “ping”, “ping”, “ping”\]  
> \[\[\], \[1\], \[100\], \[3001\], \[3002\]\]  
> Output  
> \[null, 1, 2, 3, 3\]

> Explanation  
> RecentCounter recentCounter = new RecentCounter();  
> recentCounter.ping(1); // requests = \[1\], range is \[-2999,1\], return 1  
> recentCounter.ping(100); // requests = \[1, 100\], range is \[-2900,100\], return 2  
> recentCounter.ping(3001); // requests = \[1, 100, 3001\], range is \[1,3001\], return 3  
> recentCounter.ping(3002); // requests = \[1, 100, 3001, 3002\], range is \[2,3002\], return 3

Constraints:

*  1 <= t <= 109
 *  Each test case will call ping with strictly increasing values of t.
 *  At most 104 calls will be made to ping.

# 2. 解题思路 #

## 方法1： ##

> 1.使用队列来存储最近的通话记录  
> 2.将当前通话记录加入队列  
> 3.移除时间戳小于 t - 3000 的通话记录  
> 4.返回队列中剩余通话记录的数量

# 3. 代码 #

## 代码1： ##

class RecentCounter {
        
    	Queue<Integer>que;
        public RecentCounter() {
        
        	que= new LinkedList<>();// 1.使用队列来存储最近的通话记录
        }
    
        public int ping(int t) {
        
           que.add(t);// 2.将当前通话记录加入队列
           while(que.peek()<t-3000){
         // 3.移除时间戳小于 t - 3000 的通话记录
               que.poll();
           }
           return que.size();// 4.返回队列中剩余通话记录的数量
    
         }
    }