zoukankan      html  css  js  c++  java
  • 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 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;

     [暴力解法]:

    时间分析:

    空间分析:

    [奇葩输出条件]:

    [奇葩corner case]:

    [思维问题]:

    [一句话思路]:

    直接用哈希表

    [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

    [画图]:

    [一刷]:

    没有考虑到更新的情况:0T-8F-11T-14F。 t+10k == timestamp是写不完的,不妨逆向思考 timestamp - t <10

    [二刷]:

    [三刷]:

    [四刷]:

    [五刷]:

      [五分钟肉眼debug的结果]:

    [总结]:

    不妨逆向思考 timestamp - t <10

    [复杂度]:Time complexity: O(n) Space complexity: O(n)

    [英文数据结构或算法,为什么不用别的数据结构或算法]:

    典型的key- value模式

    [关键模板化代码]:

    [其他解法]:

    [Follow Up]:

    [LC给出的题目变变变]:

    362. Design Hit Counter 五分钟以内的敲打数:没什么特色,用数据就行了吧

     [代码风格] :

    class Logger {
    
        /** Initialize your data structure here. */
        HashMap<String, Integer> map;
        
        public Logger() {
            map = new HashMap<>();
        }
        
        /** 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. */
        public boolean shouldPrintMessage(int timestamp, String message) {
            //no message
            if (!map.containsKey(message)) {
                map.put(message, timestamp);
                return true;
                 //message but time is wrong
            }else if ((timestamp - map.get(message)) >= 10) {
                map.put(message, timestamp);
                return true;
            }else {
                return false;
            } 
        }
    }
    
    /**
     * Your Logger object will be instantiated and called as such:
     * Logger obj = new Logger();
     * boolean param_1 = obj.shouldPrintMessage(timestamp,message);
     */
    View Code
  • 相关阅读:
    SQLServer中通过脚本内容查找存储过程
    TensorFlow学习笔记——节点(constant、placeholder、Variable)
    解决方案:System.InvalidOperationException: 此实现不是 Windows 平台 FIPS 验证的加密算法的一部分。
    通过网页或Serverice远程系统网站(服务)所在服务器本地的应用程序(未成功)
    (MSSQL)sp_refreshview刷新视图失败及更新Table字段失败的问题解决
    创建自己的代码片段(CodeSnippet)
    vue 创建监听,和销毁监听(addEventListener, removeEventListener)
    vue 运行时报 dependency was not found:错误
    Git 本地创建分支并提交远程分支
    vue自定义组件(通过Vue.use()来使用)即install的使用
  • 原文地址:https://www.cnblogs.com/immiao0319/p/8552190.html
Copyright © 2011-2022 走看看