边工作边刷题:70天一遍leetcode: day 97-3
Logger Rate Limiter
要点:这题并不是需要circular buffer的那题。因为没有用统计过去10秒的个数来throttle,所以不需要sliding window,而只是过去1个的timestamp来比较,所以用个hashmap记录过去的时间点即可
- 秒的间隔计算:过去10秒,意思是算上当前秒+过去9秒(当前秒和最过去的差值是9),所以-10就不在范围内了 or 条件是>=10
# 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;
class Logger(object):
def __init__(self):
"""
Initialize your data structure here.
"""
self.logtime = {}
def shouldPrintMessage(self, timestamp, message):
"""
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.
:type timestamp: int
:type message: str
:rtype: bool
"""
if message not in self.logtime:
self.logtime[message]=timestamp
return True
if timestamp - self.logtime[message]>=10:
self.logtime[message]=timestamp
return True
return False
# Your Logger object will be instantiated and called as such:
# obj = Logger()
# param_1 = obj.shouldPrintMessage(timestamp,message)