My code:

public class HitCounter {
    TreeMap<Integer, Integer> tree;
    private int counter = 0;
    /** Initialize your data structure here. */
    public HitCounter() {
        tree = new TreeMap<Integer, Integer>();
        tree.put(0, 0);
    }
    
    /** Record a hit.
        @param timestamp - The current timestamp (in seconds granularity). */
    public void hit(int timestamp) {
        counter++;
        tree.put(timestamp, counter);
    }
    
    /** Return the number of hits in the past 5 minutes.
        @param timestamp - The current timestamp (in seconds granularity). */
    public int getHits(int timestamp) {
        int end = tree.floorKey(timestamp);
        int begin = timestamp - 300;
        if (begin <= 0) {
            return tree.get(end);
        }
        else {
            begin = tree.floorKey(begin);
            return tree.get(end) - tree.get(begin);
        }
    }
}

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

自己用 TreeMap 写了出来。
然后研究了下TreeMap

他的 put, get, remove, containsKey, floor 都是 log(n) 时间复杂度。

所以这种实现方法， hit, getHits 的时间复杂度都是 O(log n)

然后看答案看到了一种新的方法：

My code:

public class HitCounter {
    Queue<Integer> q = new LinkedList<Integer>();
    /** Initialize your data structure here. */
    public HitCounter() {
        
    }
    
    /** Record a hit.
        @param timestamp - The current timestamp (in seconds granularity). */
    public void hit(int timestamp) {
        q.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 (!q.isEmpty() && timestamp - q.peek() >= 300) {
            q.poll();
        }
        
        return q.size();
    }
}

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

reference:
https://discuss.leetcode.com/topic/48752/simple-java-solution-with-explanation/2

用一个queue来做。

然后这个方法其实存在一个问题。正如题目中说的，
当hit大量出现的时候，这个方法不能scale
加入 timestamp 1-3000 都 hit
那么， getHits(3001), 就需要先弹出几千个Integer，之后再返回。。。太慢了。
而我的 Treemap 做法，则是 log n 时间内就能算出来，可以scale

Anyway, Good luck, Richardo! — 09/12/2016