[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
- 在腾讯CentOS7.4云服务器上安装Docker,在Docker上安装配置MySQL、Tomcat和Nginx
提示:以下是在腾讯CentOS7.4云服务器上操作. Docker的基本操作:https://www.cnblogs.com/opsprobe/p/10963098.html 一.安装Docker # ...
- 学习Spring-Data-Jpa(十三)---动态查询接口JpaSpecificationExecutor
1.JpaSpecificationExecutor JPA2引入了一个criteria API,我们可以使用它以编程的形式构建查询.通过编写criteria,动态生成query语句.JpaSpeci ...
- 原生js打地鼠
我们要做的是一个打地鼠的游戏,只用原生js 1.导入需要的图片 2.编写页面css样式demo.css *{ margin:0; padding:0; } .game{ position: relat ...
- 使用gitstats分析git 仓库代码
gitstats 是一个很不错的git 代码提交分析工具,可以帮助我们生成图表统计结果 工具文档信息 gitstats http://gitstats.sourceforge.net/ 安装 使用ce ...
- 洛谷 P1948 [USACO08JAN]电话线Telephone Lines 题解
P1948 [USACO08JAN]电话线Telephone Lines 题目描述 Farmer John wants to set up a telephone line at his farm. ...
- CORS跨域资源共享总结
1.CORS简述 CORS是一个W3C标准,全称是"跨域资源共享"(Cross-origin resource sharing).它允许浏览器向跨源(协议 + 域名 + 端口)服务 ...
- 微信小程序组件化开发框架WePY
wepy-CLI 安装 npm install -g wepy-cli wepy init standard my-project https://github.com/Tencent/wepy 特性 ...
- [代码质量] Maintainability Index (MI)
转载自: http://www.projectcodemeter.com/cost_estimation/help/GL_maintainability.htm ProjectCodeMeter Ma ...
- mybatis Example Criteria like 模糊查询
用Mybatis代码生成工具会产生很多个XXXExample类,这些类的作用是什么? 查阅了很多资料,在这里总结归纳一下 简介XXXExample类用于构造复杂的筛选条件 它包含一个名为Criteri ...
- 【spring源码分析】IOC容器解析
参考: https://www.iteye.com/topic/1121913(自动注入bean的扫描器) https://m.imooc.com/mip/article/34150(循环依赖的解决方 ...