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&…

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…

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…

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…

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…

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…

LeetCode:最少移动次数使得数组元素相等||[462] 题目描述 给定一个非空整数数组,找到使所有数组元素相等所需的最小移动数,其中每次移动可将选定的一个元素加1或减1. 您可以假设数组的长度最多为10000. 例如: 输入: [1,2,3] 输出: 2 说明: 只有两个动作是必要的(记得每一步仅可使其中一个元素加1或减1): [1,2,3] => [2,2,3] => [2,2,2] 题目分析 一个直观的理解是这样的,如果我们只有两个数字的话,那么我们使得他们变成相等元素的最少步数是多…

LeetCode 287. Find the Duplicate Number 暴力解法 时间 O(nlog(n)),空间O(n),按题目中Note"只用O(1)的空间",照理是过不了的,但是可能判题并没有卡空间复杂度,所以也能AC. class Solution: # 基本思路为,将第一次出现的数字 def findDuplicate(self, nums: List[int]) -> int: s = set() for i in nums: a = i in s if a…