problem 933. Number of Recent Calls 参考 1. Leetcode_easy_933. Number of Recent Calls; 完…

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 - 3…

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 - 3…

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 - 3…

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 - 3…

问题 该文章的最新版本已迁移至个人博客[比特飞],单击链接 https://www.byteflying.com/archives/4134 访问. 写一个 RecentCounter 类来计算最近的请求. 它只有一个方法:ping(int t),其中 t 代表以毫秒为单位的某个时间. 返回从 3000 毫秒前到现在的 ping 数. 任何处于 [t - 3000, t] 时间范围之内的 ping 都将会被计算在内,包括当前(指 t 时刻)的 ping. 保证每次对 ping 的调用都使用比之前…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 二分查找 队列 相似题目 参考资料 日期 题目地址:https://leetcode.com/problems/number-of-recent-calls/description/ 题目描述 Write a class RecentCounter to count recent requests. It has only one method: p…

这是悦乐书的第357次更新,第384篇原创 01 看题和准备 今天介绍的是LeetCode算法题中Easy级别的第219题(顺位题号是933).写一个类RecentCounter来计算最近的请求. 它只有一个方法:ping(int t),其中t代表一些时间(以毫秒为单位). 返回从3000毫秒前到现在为止的ping数. 在[t-3000,t]中任何时间ping都将计数,包括当前ping. 每次调用ping都使用比之前严格更大的t值.例如: 输入:inputs = ["RecentCounter&…

求最近3000毫秒内有多少次调用请求,每一次ping的时间一定比上一次的时间高:解法可以判断最后面一个数t1与最前一个数t2的差不大于3000毫秒,如果大于就直接舍弃,t1与t2之间的个数就是请求次数,可以用队列来完成. class RecentCounter { public: RecentCounter() { } int ping(int t) { Qping.push(t); while(!Qping.empty()) { ) { Qping.pop(); } else { break;…

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