[LeetCode] 362. Design Hit Counter

本文介绍了一种设计思路,用于实现一个能够记录过去5分钟内点击次数的计数器。通过两种方法实现:一是使用队列,二是使用两个数组分别存储时间和点击次数。这两种方法都能有效地在时间戳递增的情况下,统计特定时间窗口内的点击数量。

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Problem

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?

Solution #1 naive: Using Queue


class HitCounter {
    
    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) {
        if (queue.isEmpty()) return 0;
        while (!queue.isEmpty() && timestamp-queue.peek() >= 300) queue.poll();
        return queue.size();
    }
}

Solution #2 using two arrays time[] and hit[]

class HitCounter {

    int[] hit;
    int[] time;
    
    public HitCounter() {
        hit = new int[300];
        time = new int[300];
    }

    public void hit(int timestamp) {
        int index = timestamp%300;
        if (time[index] != timestamp) {
            time[index] = timestamp;
            hit[index] = 1;
        } else hit[index]++;
    }
    
    public int getHits(int timestamp) {
        int total = 0;
        for (int i = 0; i < 300; i++) {
            if (timestamp-time[i] < 300) total += hit[i];
        }
        return total;
    }
}
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