Design a logger system that receive stream of messages along with its timestamps, each message should be printed if and only if it is not printed in the last 10 seconds.

Given a message and a timestamp (in seconds granularity), return true if the message should be printed in the given timestamp, otherwise returns false.

It is possible that several messages arrive roughly at the same time.

Example:

Logger logger = new Logger();

// logging string "foo" at timestamp 1
logger.shouldPrintMessage(, "foo"); returns true; // logging string "bar" at timestamp 2
logger.shouldPrintMessage(,"bar"); returns true; // logging string "foo" at timestamp 3
logger.shouldPrintMessage(,"foo"); returns false; // logging string "bar" at timestamp 8
logger.shouldPrintMessage(,"bar"); returns false; // logging string "foo" at timestamp 10
logger.shouldPrintMessage(,"foo"); returns false; // logging string "foo" at timestamp 11
logger.shouldPrintMessage(,"foo"); returns true;

思路:用队列来记录当前的所有消息,用一个set来记录哪些消息在这个队列中。

 class Message {
public:
int timestamp;
string msg;
Message(int ts, string m) : timestamp(ts), msg(m) {}
};
class Logger {
public:
/** Initialize your data structure here. */
queue<Message> logQueue;
unordered_set<string> msgInQueue;
Logger() { } /** Returns true if the message should be printed in the given timestamp, otherwise returns false.
If this method returns false, the message will not be printed.
The timestamp is in seconds granularity. */
bool shouldPrintMessage(int timestamp, string message) {
//remove all msgs before 10s ago
while (!logQueue.empty() && logQueue.front().timestamp + <= timestamp) {
msgInQueue.erase(logQueue.front().msg);
logQueue.pop();
}
bool notInQueue = msgInQueue.count(message) == ;
//this msg will be printed, add it to log
if (notInQueue) {
logQueue.push(Message(timestamp, message));
msgInQueue.insert(message);
}
return notInQueue;
}
}; /**
* Your Logger object will be instantiated and called as such:
* Logger obj = new Logger();
* bool param_1 = obj.shouldPrintMessage(timestamp,message);
*/

Logger Rate Limiter -- LeetCode的更多相关文章

  1. 359. Logger Rate Limiter

    /* * 359. Logger Rate Limiter * 2016-7-14 by Mingyang * 很简单的HashMap,不详谈 */ class Logger { HashMap< ...

  2. [LeetCode] Logger Rate Limiter 记录速率限制器

    Design a logger system that receive stream of messages along with its timestamps, each message shoul ...

  3. LeetCode Logger Rate Limiter

    原题链接在这里:https://leetcode.com/problems/logger-rate-limiter/ 题目: Design a logger system that receive s ...

  4. LeetCode 359. Logger Rate Limiter (记录速率限制器)$

    Design a logger system that receive stream of messages along with its timestamps, each message shoul ...

  5. [LeetCode] 359. Logger Rate Limiter 记录速率限制器

    Design a logger system that receive stream of messages along with its timestamps, each message shoul ...

  6. LeetCode 359 Logger Rate Limiter

    Problem: Design a logger system that receive stream of messages along with its timestamps, each mess ...

  7. 【LeetCode】359. Logger Rate Limiter 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 ...

  8. Logger Rate Limiter 十秒限时计数器

    [抄题]: Design a logger system that receive stream of messages along with its timestamps, each message ...

  9. Logger Rate Limiter

    Design a logger system that receive stream of messages along with its timestamps, each message shoul ...

随机推荐

  1. [译]15-spring 自动装配

    前面的章节我们已经学习了如何使用bean元素在xml配置文件中声明一个bean.也学习了如何使用bean的子元素contructor-arg 和property进行bean的依赖项的注入. 之前bea ...

  2. jwt手动生成access_token

    from rest_framework_jwt.settings import api_settings # 手动为用户生成tokenjwt_payload_handler = api_setting ...

  3. 孤荷凌寒自学python第三十八天初识python的线程控制

     孤荷凌寒自学python第三十八天初识python的线程控制 (完整学习过程屏幕记录视频地址在文末,手写笔记在文末) 一.线程 在操作系统中存在着很多的可执行的应用程序,每个应用程序启动后,就可以看 ...

  4. diskimage-builder

    Supported Distributions Distributions which are supported as a build host: Centos 6, 7 Debian 8 (“je ...

  5. 原始套接字--简易ping程序

    #include <stdio.h> #include <stdlib.h> #include <string.h> #include <errno.h> ...

  6. HDU 4681 String 胡搞

    设串C的第一个字母在串A中出现的位置是stA, 串C的最后一个字母在串A中出现的位置是edA. 设串C的第一个字母在串B中出现的位置是stB, 串C的最后一个字母在串B中出现的位置是edB. 求出每一 ...

  7. ajax是可以本地运行的

    ajax是可以本地运行的,经过验证,可以是可以,但跟浏览器有关,火狐和新IE可以,chrome不可以,旧ie不知道什么原因也不可以.但是浏览器也有它的安全策略,必须是同一目录下的文件可以访问.chro ...

  8. IDEA 14注册码

    用户: share密钥:78689-AFOCD-P3SDN-83DEC-BQ3UC-V6UK7用户: for密钥:13768-8VXX0-YL2BG-YBD88-2M3HN-CAOQ5用户: you密 ...

  9. 【bzoj4819】[Sdoi2017]新生舞会 分数规划+费用流

    题目描述 学校组织了一次新生舞会,Cathy作为经验丰富的老学姐,负责为同学们安排舞伴.有n个男生和n个女生参加舞会买一个男生和一个女生一起跳舞,互为舞伴.Cathy收集了这些同学之间的关系,比如两个 ...

  10. LeetCode -- Product of Array Except Self My Submissions Question

    Question: Given an array of n integers where n > 1, nums, return an array output such that output ...