Design a logger system that receives a stream of messages along with their timestamps. Each unique message should only be printed at most every 10 seconds (i.e. a message printed at timestamp
t
will prevent other identical messages from being printed until timestamp t + 10
).All messages will come in chronological order. Several messages may arrive at the same timestamp.
Implement the
Logger
class:Logger()
Initializes the logger
object.
bool shouldPrintMessage(int timestamp, string message)
Returns true
if the message
should be printed in the given timestamp
, otherwise returns false
.Example 1:
Input
[“Logger”, “shouldPrintMessage”, “shouldPrintMessage”, “shouldPrintMessage”, “shouldPrintMessage”, “shouldPrintMessage”, “shouldPrintMessage”]
[[], [1, “foo”], [2, “bar”], [3, “foo”], [8, “bar”], [10, “foo”], [11, “foo”]]
Output
[null, true, true, false, false, false, true]
Explanation
Logger logger = new Logger();
logger.shouldPrintMessage(1, “foo”); // return true, next allowed timestamp for “foo” is 1 + 10 = 11
logger.shouldPrintMessage(2, “bar”); // return true, next allowed timestamp for “bar” is 2 + 10 = 12
logger.shouldPrintMessage(3, “foo”); // 3 < 11, return false
logger.shouldPrintMessage(8, “bar”); // 8 < 12, return false
logger.shouldPrintMessage(10, “foo”); // 10 < 11, return false
logger.shouldPrintMessage(11, “foo”); // 11 >= 11, return true, next allowed timestamp for “foo” is 11 + 10 = 21
Constraints:
0 <= timestamp <= 10<sup>9</sup>
Every timestamp
will be passed in non-decreasing order (chronological order).1 <= message.length <= 30
At most 10<sup>4</sup>
calls will be made to shouldPrintMessage
.题目大意:
实现Logger打印的rate limiter
解题思路:
题不难,但有实际意义
解题步骤:
N/A
注意事项:
- shouldPrintMessage只有返回True时候才记录时间点。否则不记录。这属于元素相等的test case
Python代码:
1 | class Logger(TestCases): |
算法分析:
时间复杂度为O(1)
,空间复杂度O(n)