362. 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); 

一刷
题解：用两个数组times, hits，长度均为300(5分钟），每次index = timetamp % 300,
如果times[index]!=timestamp, hits[index]需要重新计数。每次getSum的时候，将hit数组累加一下，如果早于5分钟前，不加进去。

class HitCounter {
    private int[] times;
    private int[] hits;

    /** Initialize your data structure here. */
    public HitCounter() {
        times = new int[300];
        hits = new int[300];
    }
    
    /** Record a hit.
        @param timestamp - The current timestamp (in seconds granularity). */
    public void hit(int timestamp) {
        int index = timestamp%300;
        if(times[index]!=timestamp){
            times[index] = timestamp;
            hits[index] = 1;
        }else hits[index]++;
    }
    
    /** Return the number of hits in the past 5 minutes.
        @param timestamp - The current timestamp (in seconds granularity). */
    public int getHits(int timestamp) {
        int total = 0;
        for(int i=0; i<300; i++){
            if(timestamp-times[i]<300) total+=hits[i];
        }
        return total;
    }
}

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