359. Logger Rate Limiter

最新推荐文章于 2025-01-24 12:24:31 发布

zshouyi

最新推荐文章于 2025-01-24 12:24:31 发布

阅读量446

点赞数

分类专栏： LeetCode Hash Table Design 文章标签： Hash Table Design Leetcode Logger Rate Limiter

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/54174281

版权

LeetCode 同时被 3 个专栏收录

402 篇文章

订阅专栏

Hash Table

51 篇文章

订阅专栏

Design

23 篇文章

订阅专栏

本文介绍了一种消息日志系统的设计，该系统确保相同消息的打印间隔至少为10秒，通过使用哈希映射记录消息及其最近打印时间来实现这一目标。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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;

Credits:
Special thanks to @memoryless for adding this problem and creating all test cases.

这道题题目很长，其实就是给个时间戳和消息，每个消息的log间隔必须大于10秒。思路是先判断有没有，没有说明第一次，就把这个string和timestamp存下来，返回true，如果有，判断上一次的时间戳是否间隔10秒，有的话哽你时间戳，返回true，间隔不足10秒的话返回false。代码如下：

public class Logger {
    HashMap<String, Integer> hm = new HashMap<String, Integer>();

    /** Initialize your data structure here. */
    public 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. */
    public boolean shouldPrintMessage(int timestamp, String message) {
        if (hm.containsKey(message)) {
            if (timestamp - hm.get(message) >= 10) {
                hm.put(message, timestamp);
                return true;
            } else {
                return false;
            }
        } else {
            hm.put(message, timestamp);
            return true;
        }
    }
}

/**
 * Your Logger object will be instantiated and called as such:
 * Logger obj = new Logger();
 * boolean param_1 = obj.shouldPrintMessage(timestamp,message);
 */

上面的思路代码有些长，换种思路，首先判断时间间隔够不够10秒，不够返回false，够的话存当前时间戳加10秒，返回true。当string不存在时，用到Map.getOrDefault()函数返回0值，肯定小于当前时间戳，跳到存储和返回true。代码如下：

public class Logger {

    private Map<String, Integer> ok = new HashMap<>();

    public boolean shouldPrintMessage(int timestamp, String message) {
        if (timestamp < ok.getOrDefault(message, 0))
            return false;
        ok.put(message, timestamp + 10);
        return true;
    }
}