359.-logger-rate-limiter.md

359. Logger Rate Limiter

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;

题目大意:输入一串附带时间戳(int)的信息(string),判断这个信息是否在十个时间戳以内输出过,若没有,则返回true,否则false 解题思路:存入一个map< string, int>中,每次比对时间戳与map中int的差值,看是否大于10即可.

class Logger {
public:
    /** Initialize your data structure here. */
    Logger() {}

    bool shouldPrintMessage(int timestamp, string message) {
        auto search = log.find(message);  //查找
        if (search != log.end()) {  //若找到
            if (abs(search->second - timestamp) >= 10) {  //看时间戳
                log[message] = timestamp;
                return true;
            } else {  
                return false;
            }
        } else {  //若找不到
            log[message] = timestamp;
            return true;
        }
    }
private:
    map<string, int> log;  //存储map
};