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题目
Write a class RecentCounter to count recent requests.
It has only one method: ping(int t), where t represents some time in milliseconds.
Return the number of pings that have been made from 3000 milliseconds ago until now.
Any ping with time in [t – 3000, t] will count, including the current ping.
It is guaranteed that every call to ping uses a strictly larger value of t than before.
Example 1:
Input: inputs = [“RecentCounter”,”ping”,”ping”,”ping”,”ping”], inputs = [[],[1],[100],[3001],[3002]]
Output: [null,1,2,3,3]
Note:
Each test case will have at most 10000 calls to ping.
Each test case will call ping with strictly increasing values of t.
Each call to ping will have 1 <= t <= 10^9.
题目地址
讲解
这道题最重要的是看懂题，每个 ping 都附带一个时间点，求这个时间点往前 3000 毫秒内的 ping 数量。这里我使用了一个队列来实现。
Java 代码
class RecentCounter {
private Queue<Integer> queue;
public RecentCounter() {
queue = new LinkedList<>();
}
public int ping(int t) {
while(queue.size()>0 && queue.peek()<t-3000){
queue.poll();
}
queue.offer(t);
return queue.size();
}
}
/**
* Your RecentCounter object will be instantiated and called as such:
* RecentCounter obj = new RecentCounter();
* int param_1 = obj.ping(t);
*/
