# 933. Number of Recent Calls

Write a class `RecentCounter` to count recent requests.

It has only one method: `ping(int t)`, where t represents some time in milliseconds.

Return the number of `ping`s that have been made from 3000 milliseconds ago until now.

Any ping with time in `[t - 3000, t]` will count, including the current ping.

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

Example 1:

``````Input: inputs = ["RecentCounter","ping","ping","ping","ping"], inputs = [[],[1],[100],[3001],[3002]]
Output: [null,1,2,3,3]
``````

Note:

1. Each test case will have at most `10000` calls to `ping`.

2. Each test case will call `ping` with strictly increasing values of `t`.

3. Each call to ping will have `1 <= t <= 10^9`.

这道题让实现一个 RecentCounter 类，里面有一个 ping 函数，输入给定了一个时间t，让我们求在 [t-3000, t] 时间范围内有多少次 ping。题目中限定了每次的给的时间一定会比上一次的时间大，而且只关心这个大小为 3001 的时间窗口范围内的次数，则利用滑动窗口 Sliding Window 来做就是个很不错的选择。由于数字是不断加入的，可以使用一个 queue，每当要加入一个新的时间点t时，先从队列开头遍历，若前面的时间不在当前的时间窗口内，则移除队列。之后再将当前时间点t加入，并返回队列的长度即可，参见代码如下：

class RecentCounter {
public:
RecentCounter() {}

int ping(int t) {

``````   while (!q.empty()) {
if (q.front() + 3000 >= t) break;
q.pop();
}
q.push(t);
return q.size();
``````

}

private:
queue q;
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/933

https://leetcode.com/problems/number-of-recent-calls/

https://leetcode.com/problems/number-of-recent-calls/discuss/189334/C%2B%2B-Easy-and-Clean-solution-using-queue

https://leetcode.com/problems/number-of-recent-calls/discuss/189239/JavaPython-3-Five-solutions%3A-TreeMap-TreeSet-ArrayList-Queue-Circular-List.

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