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?pings 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:
- Each test case will have at most?
10000?calls to?ping. - Each test case will call?
ping?with strictly increasing values of?t. - 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<int> q;
};Github 同步地址:
https://github.com/grandyang/leetcode/issues/933
参考资料：
https://leetcode.com/problems/number-of-recent-calls/
LeetCode All in One 题目讲解汇总(持续更新中...)
原文地址：https://www.cnblogs.com/grandyang/p/12359774.html