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  • LeetCode 359. 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 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;

    题解:

    再次看到这题就像回到了当年的战场.

    维护一个HashMap, key 是message, value 是该message的timestamp

    没出现过的message加到HashMap中, return true.

    出现过并没有超过10的message return false.

    出现过并超过10的message, 跟新HashMap, return true.

    Time Complexity: shouldPrintMessage O(1).

    Space: HashMap的大小.

    AC Java:

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

     跟上Design Hit Counter.

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  • 原文地址:https://www.cnblogs.com/Dylan-Java-NYC/p/6197022.html
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