[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
- 微信之获取微信的openid(二)详细版
第一步 :配置测试号,网页授权获取用户基本信息. 该授权回掉页面域名为ngrok 映射的域名,我的映射地址是127.0.0.1:8080. 到此微信配置完毕,接下来就是直接上代码了 2.用户同意授权 ...
- background-image:url为空引发的两次请求问题
参考文章: https://blog.csdn.net/jsjhushilei/article/details/51101014 1.Nicholas 在 2009 年就开始推动各浏览器厂商,现在看起 ...
- destoon下动态链接301到伪静态(ngnix)
分享一个destoon6.0/7.0下动态链接301到伪静态上面,实现权重提升. if ($request_uri ~* "^/index.php\?itemid=(\d+)&mod ...
- flume的sink写入hive表
flume的配置文件如下: a1.sources=r1 a1.channels=c1 a1.sinks=s1 a1.sources.r1.type=netcat a1.sources.r1.bind= ...
- linux中的alias命令详解
功能说明:设置指令的别名.语 法:alias[别名]=[指令名称]参 数 :若不加任何参数,则列出目前所有的别名设置.举 例 :ermao@lost-desktop:~$ alias ...
- django-mysql事务
django文档:https://yiyibooks.cn/xx/django_182/topics/db/transactions.html mysql事务 1) 事务概念 一组mysql语句,要么 ...
- ARDUNIO IMU processing姿态数据可视化
https://www.arduino.cn/thread-42852-1-1.html 关键数据打包 float roll, pitch, heading; Serial.print("O ...
- LOJ P10013 曲线 题解
每日一题 day38 打卡 Analysis 这道题运用的是三分,就是说具有一定的单调性,找最大最小值,然后和二分基本类似,就是说特性就是说当前两个点比较,较优的点和最优点在相对了较差点的同侧,就是说 ...
- 持续集成学习9 jenkins执行脚本
一.配置 1.首先在slave节点上写一脚本 [root@node1 script]# cat /application/script/test.sh #!/bin/bash echo "h ...
- 手工部署yugabyte的几点说明
ntp 时间同步 ntp 时间同步对于yugabyte 是一个比较重要的服务,需要注意时间的同步 YB-Master 个数的说明 原则 YB-Master 的个数,必须和复制因子的个数一样,同时mas ...