0933. Number of Recent Calls (E)

Number of Recent Calls (E)

题目

You have a RecentCounter class which counts the number of recent requests within a certain time frame.

Implement the RecentCounter class:

• RecentCounter() Initializes the counter with zero recent requests.
• int ping(int t) Adds a new request at time t, where t represents some time in milliseconds, and returns the number of requests that has happened in the past 3000 milliseconds (including the new request). Specifically, return the number of requests that have happened in the inclusive range [t - 3000, t].

It is guaranteed that every call to ping uses a strictly larger value of t than the previous call.

Example 1:

Input
["RecentCounter", "ping", "ping", "ping", "ping"]
[[], [1], [100], [3001], [3002]]
Output
[null, 1, 2, 3, 3]

Explanation
RecentCounter recentCounter = new RecentCounter();
recentCounter.ping(1);     // requests = [1], range is [-2999,1], return 1
recentCounter.ping(100);   // requests = [1, 100], range is [-2900,100], return 2
recentCounter.ping(3001);  // requests = [1, 100, 3001], range is [1,3001], return 3
recentCounter.ping(3002);  // requests = [1, 100, 3001, 3002], range is [2,3002], return 3

Constraints:

• 1 <= t <= 104
• Each test case will call ping with strictly increasing values of t.
• At most 104 calls will be made to ping.

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题意

在t时间增长一条记录，同时返回[t-3000, t]时间段内记录的条数。

思路

两种方法：

1. 保存全部记录，用二分法找到对应t-3000的位置；
2. 利用滑动窗口，每次加入一条新记录后，删去全部t-3000以前的记录。

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代码实现

Java

二分法

class RecentCounter {
    private List<Integer> record;

    public RecentCounter() {
        record = new ArrayList<>();
    }

    public int ping(int t) {
        record.add(t);
        return record.size() - find(t - 3000);
    }

    private int find(int target) {
        int left = 0, right = record.size() - 1;
        while (left < right) {
            int mid = (right - left) / 2 + left;
            if (record.get(mid) >= target) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }
}

滑动窗口

class RecentCounter {
    private Deque<Integer> record;

    public RecentCounter() {
        record = new LinkedList<>();
    }

    public int ping(int t) {
        record.offer(t);
        while (record.getFirst() < t - 3000) {
            record.poll();
        }
        return record.size();
    }
}