[LeetCode] 362. Design Hit Counter 设计点击计数器
Design a hit counter which counts the number of hits received in the past 5 minutes.
Each function accepts a timestamp parameter (in seconds granularity) and you may assume that calls are being made to the system in chronological order (ie, the timestamp is monotonically increasing). You may assume that the earliest timestamp starts at 1.
It is possible that several hits arrive roughly at the same time.
Example:
HitCounter counter = new HitCounter(); // hit at timestamp 1.
counter.hit(1); // hit at timestamp 2.
counter.hit(2); // hit at timestamp 3.
counter.hit(3); // get hits at timestamp 4, should return 3.
counter.getHits(4); // hit at timestamp 300.
counter.hit(300); // get hits at timestamp 300, should return 4.
counter.getHits(300); // get hits at timestamp 301, should return 3.
counter.getHits(301);
Follow up:
What if the number of hits per second could be very large? Does your design scale?
设计一个点击计数器,能够返回五分钟内的点击数,提示了有可能同一时间内有多次点击。
解法:要求保证时间顺序,可以用一个queue将每次点击的timestamp放入queue中。getHits: 可以从queue的头开始看, 如果queue开头的时间在范围外,就poll掉。最后返回queue的size。
Java:
public class HitCounter {
private ArrayDeque<Integer> queue;
/** Initialize your data structure here. */
public HitCounter() {
queue = new ArrayDeque<Integer>();
}
/** Record a hit.
@param timestamp - The current timestamp (in seconds granularity). */
public void hit(int timestamp) {
queue.offer(timestamp);
}
/** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity). */
public int getHits(int timestamp) {
int startTime = timestamp - 300;
while(!queue.isEmpty() && queue.peek() <= startTime) {
queue.poll();
}
return queue.size();
}
}
/**
* Your HitCounter object will be instantiated and called as such:
* HitCounter obj = new HitCounter();
* obj.hit(timestamp);
* int param_2 = obj.getHits(timestamp);
*/
Java:
public class HitCounter {
private Hit start = new Hit(0);
private Hit tail = start;
private int count = 0;
/** Initialize your data structure here. */
public HitCounter() {
}
/** Record a hit.
@param timestamp - The current timestamp (in seconds granularity). */
public void hit(int timestamp) {
if (tail.timestamp == timestamp) {
tail.count ++;
count ++;
} else {
tail.next = new Hit(timestamp);
tail = tail.next;
count ++;
}
getHits(timestamp);
}
/** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity). */
public int getHits(int timestamp) {
while (start.next != null && timestamp - start.next.timestamp >= 300) {
count -= start.next.count;
start.next = start.next.next;
}
if (start.next == null) tail = start;
return count;
}
}
class Hit {
int timestamp;
int count;
Hit next;
Hit(int timestamp) {
this.timestamp = timestamp;
this.count = 1;
}
}
/**
* Your HitCounter object will be instantiated and called as such:
* HitCounter obj = new HitCounter();
* obj.hit(timestamp);
* int param_2 = obj.getHits(timestamp);
*/
Python:
# Time: O(1), amortized
# Space: O(k), k is the count of seconds. from collections import deque class HitCounter(object): def __init__(self):
"""
Initialize your data structure here.
"""
self.__k = 300
self.__dq = deque()
self.__count = 0 def hit(self, timestamp):
"""
Record a hit.
@param timestamp - The current timestamp (in seconds granularity).
:type timestamp: int
:rtype: void
"""
self.getHits(timestamp)
if self.__dq and self.__dq[-1][0] == timestamp:
self.__dq[-1][1] += 1
else:
self.__dq.append([timestamp, 1])
self.__count += 1 def getHits(self, timestamp):
"""
Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity).
:type timestamp: int
:rtype: int
"""
while self.__dq and self.__dq[0][0] <= timestamp - self.__k:
self.__count -= self.__dq.popleft()[1]
return self.__count # Your HitCounter object will be instantiated and called as such:
# obj = HitCounter()
# obj.hit(timestamp)
# param_2 = obj.getHits(timestamp)
C++:
// Time: O(1), amortized
// Space: O(k), k is the count of seconds. class HitCounter {
public:
/** Initialize your data structure here. */
HitCounter() : count_(0) { } /** Record a hit.
@param timestamp - The current timestamp (in seconds granularity). */
void hit(int timestamp) {
getHits(timestamp);
if (!dq_.empty() && dq_.back().first == timestamp) {
++dq_.back().second;
} else {
dq_.emplace_back(timestamp, 1);
}
++count_;
} /** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity). */
int getHits(int timestamp) {
while (!dq_.empty() && dq_.front().first <= timestamp - k_) {
count_ -= dq_.front().second;
dq_.pop_front();
}
return count_;
} private:
const int k_ = 300;
int count_;
deque<pair<int, int>> dq_;
}; /**
* Your HitCounter object will be instantiated and called as such:
* HitCounter obj = new HitCounter();
* obj.hit(timestamp);
* int param_2 = obj.getHits(timestamp);
*/
C++:
class HitCounter {
public:
/** Initialize your data structure here. */
HitCounter() {}
/** Record a hit.
@param timestamp - The current timestamp (in seconds granularity). */
void hit(int timestamp) {
q.push(timestamp);
}
/** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity). */
int getHits(int timestamp) {
while (!q.empty() && timestamp - q.front() >= 300) {
q.pop();
}
return q.size();
}
private:
queue<int> q;
};
C++:
class HitCounter {
public:
/** Initialize your data structure here. */
HitCounter() {}
/** Record a hit.
@param timestamp - The current timestamp (in seconds granularity). */
void hit(int timestamp) {
v.push_back(timestamp);
}
/** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity). */
int getHits(int timestamp) {
int i, j;
for (i = 0; i < v.size(); ++i) {
if (v[i] > timestamp - 300) {
break;
}
}
return v.size() - i;
}
private:
vector<int> v;
};
C++:
class HitCounter {
public:
/** Initialize your data structure here. */
HitCounter() {
times.resize(300);
hits.resize(300);
}
/** Record a hit.
@param timestamp - The current timestamp (in seconds granularity). */
void hit(int timestamp) {
int idx = timestamp % 300;
if (times[idx] != timestamp) {
times[idx] = timestamp;
hits[idx] = 1;
} else {
++hits[idx];
}
}
/** Return the number of hits in the past 5 minutes.
@param timestamp - The current timestamp (in seconds granularity). */
int getHits(int timestamp) {
int res = 0;
for (int i = 0; i < 300; ++i) {
if (timestamp - times[i] < 300) {
res += hits[i];
}
}
return res;
}
private:
vector<int> times, hits;
};
类似题目:
[LeetCode] 359. Logger Rate Limiter 记录速率限制器
All LeetCode Questions List 题目汇总
[LeetCode] 362. Design Hit Counter 设计点击计数器的更多相关文章
- [LeetCode] Design Hit Counter 设计点击计数器
Design a hit counter which counts the number of hits received in the past 5 minutes. Each function a ...
- LeetCode 362. Design Hit Counter
原题链接在这里:https://leetcode.com/problems/design-hit-counter/description/ 题目: Design a hit counter which ...
- 【LeetCode】362. Design Hit Counter 解题报告 (C++)
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 字典 日期 题目地址:https://leetcode ...
- [LC] 362. Design Hit Counter
Design a hit counter which counts the number of hits received in the past 5 minutes. Each function a ...
- 362. Design Hit Counter
这个傻逼题..我没弄明白 you may assume that calls are being made to the system in chronological order (ie, the ...
- [LeetCode] 641.Design Circular Deque 设计环形双向队列
Design your implementation of the circular double-ended queue (deque). Your implementation should su ...
- [LeetCode] 622.Design Circular Queue 设计环形队列
Design your implementation of the circular queue. The circular queue is a linear data structure in w ...
- LeetCode Design Hit Counter
原题链接在这里:https://leetcode.com/problems/design-hit-counter/. 题目: Design a hit counter which counts the ...
- Design Hit Counter
Design a hit counter which counts the number of hits received in the past 5 minutes. Each function a ...
随机推荐
- python代码规范 自动优化工具Black
自动优化工具Black 在众多代码格式化工具中,Black算是比较新的一个,它***的特点是可配置项比较少,个人认为这对于新手来说是件好事,因为我们不必过多考虑如何设置Black,让 Black 自己 ...
- Python使用pip安装matplotlib模块
matplotlib是python中强大的画图模块. 首先确保已经安装python,然后用pip来安装matplotlib模块. 进入到cmd窗口下,建议执行python -m pip install ...
- 牛客1024B 石头游戏
题目描述 石头游戏在一个 \(n\) 行 \(m\) 列 \((1\leq n,m \leq 8)(1≤n,m≤8)\) 的网格上进行,每个格子对应一种操作序列,操作序列至多有10种,分别用0~9这1 ...
- Intellij IDEA 与 Gitlab 实现代码上传与下载
整体流程:1.download project2.deposit project structure and set default server3.configure tomcat 2 steps4 ...
- postgresql从库提升为主库
一.停主库 1.查看当前连接 select pid,datname,usename,client_addr,client_port, application_name from pg_stat_act ...
- super()函数
1.简单的使用 在类的继承中,如果重定义某个方法,该方法会覆盖父类的同名方法,但有时,我们希望能同时实现父类的功能,这时,我们就需要调用父类的方法了,可通过使用 super 来实现,比如: 在上面,A ...
- 使用SpringBoot访问jsp页面
1 编写application.yml文件 spring: mvc: view: suffix: .jsp prefix: /jsp/ 2 创建Controller层 @Controller @Req ...
- MongoDB 4.2 的主要亮点(转载)
在6月份召开的MongoDB全球用户大会上, MongoDB官宣了MongoDB Server 4.2,在经过100,000多个运行实例的测试后,MongoDB 4.2表现强劲.现在4.2版本正式上线 ...
- BZOJ 4771: 七彩树 可持久化线段树+树链的并
这个思路挺有意思的 ~ 利用树链的并来保证每个颜色只贡献一次,然后用可持久化线段树维护 code: #include <set> #include <cstdio> #incl ...
- python(三)——while语句
while死循环 #!/usr/bin/env python #-*- coding:utf8 -*- import time while 1 == 1: print('Ok',time.time() ...