Design Hit Counter

https://leetcode.com/problems/design-hit-counter/description/

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?

Thoughts

题目意思容易产生歧义. 首先时间是模拟现实时间, 一直是递增的, getHits的时间是"当前的时间", 也不会比之前的任何hit时间小. 要求只维持300秒内的hits数即可. 因此可以用个queue把超出300s的太老的挤出去. 这个思想也可以用数组配合mod来实现.

Code

class HitCounter {
    private Queue<Integer> queue;

    /** Initialize your data structure here. */
    public HitCounter() {
        queue = new LinkedList<>();
    }

    /** Record a hit.
        @param timestamp - The current timestamp (in seconds granularity). */
    public void hit(int timestamp) {
        queue.offer(timestamp);
    }

    /** Return the number of hits in the past 5 minutes.
        @param timestamp - The current timestamp (in seconds granularity). */
    public int getHits(int timestamp) {
        while (!queue.isEmpty() && timestamp - queue.peek() >= 300) {
            queue.poll();
        }
        return queue.size();
    }
}

/**
 * Your HitCounter object will be instantiated and called as such:
 * HitCounter obj = new HitCounter();
 * obj.hit(timestamp);
 * int param_2 = obj.getHits(timestamp);
 */

Analysis

时间复杂度平均O(1).