Design Hit Counter

Design a hit counter which counts the number of hits received in the past 5 minutes.

Each function accepts a timestamp parameter (in seconds granularity) and you may assume that calls are being made to the system in chronological order (ie, the timestamp is monotonically increasing). You may assume that the earliest timestamp starts at 1.

It is possible that several hits arrive roughly at the same time.

Example:

HitCounter counter = new HitCounter();

// hit at timestamp 1.
counter.hit(1);

// hit at timestamp 2.
counter.hit(2);

// hit at timestamp 3.
counter.hit(3);

// get hits at timestamp 4, should return 3.
counter.getHits(4);

// hit at timestamp 300.
counter.hit(300);

// get hits at timestamp 300, should return 4.
counter.getHits(300);

// get hits at timestamp 301, should return 3.
counter.getHits(301); 

Follow up: What if the number of hits per second could be very large? Does your design scale?

My Solutions:

设计一个点击计数器，能够返回5分钟内的点击数，同一时间有可能有多个点击。因为操作是按时间顺序的，可以用queue记录时间戳。在需要获取点击数的时候，poll出超过5分钟的时间戳，返回queue里元素的个数。

hit: Time: O(1)

getHits: Time: O(n); Sapce: O(n)

public class HitCounter {
    private Queue<Integer> q;
    public HitCounter() {
        q = new LinkedList<>();
    }
    
    public void hit(int timestamp) {
        q.offer(timestamp);
    }
    
    public int getHits(int timestamp) {
        while (!q.isEmpty() && q.peek() <= timestamp - 300) q.poll();
        return q.size();
    }
}

如果每秒会有很多点击，定义两个大小为300秒的数组times和hits，分别用来保存时间戳和点击数。在点击函数中，将时间戳对300取余，然后看此位置中之前保存的时间戳和当前的时间戳是否一样。一样说明是同一个时间戳，那么对应的点击数自增1；如果不一样，说明已经过了五分钟了，那么将对应的点击数重置为1。在返回点击数时，需要遍历times数组，找出所有在5分中内的位置，然后把hits中对应位置的点击数都加起来即可。

    public void hit(int timestamp) {
        int idx = timestamp % 300;
        if (times[idx] != timestamp) { // 重置
            times[idx] = timestamp;
            hits[idx] = 1;
        } else {
            hits[idx]++;
        }
    }

    public int getHits(int timestamp) {
        int res = 0;
        for (int i = 0; i < 300; i++) {
            if (timestamp - times[i] < 300) {
                res += hits[i];
            }
        }
        return res;
    }