zoukankan      html  css  js  c++  java
  • 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 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(1, "foo"); returns true; 
    
    // logging string "bar" at timestamp 2
    logger.shouldPrintMessage(2,"bar"); returns true;
    
    // logging string "foo" at timestamp 3
    logger.shouldPrintMessage(3,"foo"); returns false;
    
    // logging string "bar" at timestamp 8
    logger.shouldPrintMessage(8,"bar"); returns false;
    
    // logging string "foo" at timestamp 10
    logger.shouldPrintMessage(10,"foo"); returns false;
    
    // logging string "foo" at timestamp 11
    logger.shouldPrintMessage(11,"foo"); returns true;
    

    题目标签:Hash Table

      题目让我们设计一个 message 是否可以发送的检测器,只有在这个message 第一次发送,或者 上一次发送已经过去了10秒的情况下,才可以发送。

      利用HashMap 来保存 message, 和它的 timestamp 就可以了。

    Java Solution:

    Runtime beats 49.91% 

    完成日期:06/06/2017

    关键词:HashMap

    关键点:保存message 和 timestamp

     1 class Logger 
     2 {
     3     HashMap<String, Integer> map;
     4     /** Initialize your data structure here. */
     5     public Logger() 
     6     {
     7         map = new HashMap<>();
     8     }
     9     
    10     /** Returns true if the message should be printed in the given timestamp, otherwise returns false.
    11         If this method returns false, the message will not be printed.
    12         The timestamp is in seconds granularity. */
    13     public boolean shouldPrintMessage(int timestamp, String message) 
    14     {
    15         if(!map.containsKey(message)) // first time for this message
    16         {
    17             map.put(message, timestamp);
    18             return true;
    19         }
    20         else // at least second time for this message
    21         {
    22             if(timestamp >= map.get(message) + 10)
    23             {
    24                 map.put(message, timestamp);
    25                 return true;
    26             }
    27             else
    28                 return false;
    29             
    30         }
    31     }
    32 }
    33 
    34 /**
    35  * Your Logger object will be instantiated and called as such:
    36  * Logger obj = new Logger();
    37  * boolean param_1 = obj.shouldPrintMessage(timestamp,message);
    38  */

    参考资料:N/A

    LeetCode 题目列表 - LeetCode Questions List

  • 相关阅读:
    前端编程规范记录
    搬砖工坑爹教程
    JS的模块化编程学习之旅
    后端开发遇到的问题
    git学习中遇到的疑难杂症
    微信小程序填坑之旅
    详解Redis中两种持久化机制RDB和AOF
    redis系列:RDB持久化与AOF持久化
    Java中判断字符串是否为数字
    @Aspect 注解使用详解
  • 原文地址:https://www.cnblogs.com/jimmycheng/p/7797169.html
Copyright © 2011-2022 走看看