359. 日志速率限制器
请你设计一个日志系统，可以流式接收消息以及它的时间戳。每条 不重复 的消息最多只能每 10 秒打印一次。也就是说，如果在时间戳 t 打印某条消息，那么相同内容的消息直到时间戳变为 t + 10 之前都不会被打印。
所有消息都按时间顺序发送。多条消息可能到达同一时间戳。
实现 Logger 类：
Logger() 初始化 logger 对象bool shouldPrintMessage(int timestamp, string message) 如果这条消息 message 在给定的时间戳 timestamp 应该被打印出来，则返回 true ，否则请返回 false 。
示例：
输入： ["Logger", "shouldPrintMessage", "shouldPrintMessage", "shouldPrintMessage", "shouldPrintMessage", "shouldPrintMessage", "shouldPrintMessage"] [[], [1, "foo"], [2, "bar"], [3, "foo"], [8, "bar"], [10, "foo"], [11, "foo"]] 输出： [null, true, true, false, false, false, true] 解释： Logger logger = new Logger(); logger.shouldPrintMessage(1, "foo"); // 返回 true ，下一次 "foo" 可以打印的时间戳是 1 + 10 = 11 logger.shouldPrintMessage(2, "bar"); // 返回 true ，下一次 "bar" 可以打印的时间戳是 2 + 10 = 12 logger.shouldPrintMessage(3, "foo"); // 3 < 11 ，返回 false logger.shouldPrintMessage(8, "bar"); // 8 < 12 ，返回 false logger.shouldPrintMessage(10, "foo"); // 10 < 11 ，返回 false logger.shouldPrintMessage(11, "foo"); // 11 >= 11 ，返回 true ，下一次 "foo" 可以打印的时间戳是 11 + 10 = 21
提示：
0 <= timestamp <= 109timestamp 都将按非递减顺序（时间顺序）传递1 <= message.length <= 30104 次 shouldPrintMessage 方法相似题目
原站题解
golang 解法, 执行用时: 48 ms, 内存消耗: 7.9 MB, 提交时间: 2023-10-15 18:34:20
type Logger struct {
m map[string] int
}
/** Initialize your data structure here. */
func Constructor() Logger {
return Logger{m:map[string]int{}}
}
/** 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. */
func (this *Logger) ShouldPrintMessage(timestamp int, message string) bool {
res := false
if ts, ok := this.m[message]; !ok{
res = true
this.m[message] = timestamp
}else{
if timestamp - ts >= 10{
res = true
this.m[message] = timestamp
}
}
return res
}
/**
* Your Logger object will be instantiated and called as such:
* obj := Constructor();
* param_1 := obj.ShouldPrintMessage(timestamp,message);
*/
python3 解法, 执行用时: 124 ms, 内存消耗: 23 MB, 提交时间: 2023-10-15 18:33:51
from collections import deque
class Logger:
def __init__(self):
"""
Initialize your data structure here.
"""
self._msg_set = set()
self._msg_queue = deque()
def shouldPrintMessage(self, timestamp, message):
"""
Returns true if the message should be printed in the given timestamp, otherwise returns false.
"""
while self._msg_queue:
msg, ts = self._msg_queue[0]
if timestamp - ts >= 10:
self._msg_queue.popleft()
self._msg_set.remove(msg)
else:
break
if message not in self._msg_set:
self._msg_set.add(message)
self._msg_queue.append((message, timestamp))
return True
else:
return False
# Your Logger object will be instantiated and called as such:
# obj = Logger()
# param_1 = obj.shouldPrintMessage(timestamp,message)
python3 解法, 执行用时: 112 ms, 内存消耗: 23 MB, 提交时间: 2023-10-15 18:33:31
class Logger:
def __init__(self):
"""
Initialize your data structure here.
"""
self._msg_dict = {}
def shouldPrintMessage(self, timestamp, message):
"""
Returns true if the message should be printed in the given timestamp, otherwise returns false.
"""
if message not in self._msg_dict:
# case 1). add the message to print
self._msg_dict[message] = timestamp
return True
if timestamp - self._msg_dict[message] >= 10:
# case 2). update the timestamp of the message
self._msg_dict[message] = timestamp
return True
else:
return False
# Your Logger object will be instantiated and called as such:
# obj = Logger()
# param_1 = obj.shouldPrintMessage(timestamp,message)
java 解法, 执行用时: 29 ms, 内存消耗: 53.2 MB, 提交时间: 2023-10-15 18:33:07
class Logger {
private HashMap<String, Integer> msgDict;
/** Initialize your data structure here. */
public Logger() {
msgDict = new HashMap<String, Integer>();
}
/**
* Returns true if the message should be printed in the given timestamp, otherwise returns false.
*/
public boolean shouldPrintMessage(int timestamp, String message) {
if (!this.msgDict.containsKey(message)) {
this.msgDict.put(message, timestamp);
return true;
}
Integer oldTimestamp = this.msgDict.get(message);
if (timestamp - oldTimestamp >= 10) {
this.msgDict.put(message, timestamp);
return true;
} else
return false;
}
}
/**
* Your Logger object will be instantiated and called as such:
* Logger obj = new Logger();
* boolean param_1 = obj.shouldPrintMessage(timestamp,message);
*/
java 解法, 执行用时: 28 ms, 内存消耗: 53.2 MB, 提交时间: 2023-10-15 18:32:51
class Pair<U, V> {
public U first;
public V second;
public Pair(U first, V second) {
this.first = first;
this.second = second;
}
}
class Logger {
private LinkedList<Pair<String, Integer>> msgQueue;
private HashSet<String> msgSet;
/** Initialize your data structure here. */
public Logger() {
msgQueue = new LinkedList<Pair<String, Integer>>();
msgSet = new HashSet<String>();
}
/**
* Returns true if the message should be printed in the given timestamp, otherwise returns false.
*/
public boolean shouldPrintMessage(int timestamp, String message) {
// clean up.
while (msgQueue.size() > 0) {
Pair<String, Integer> head = msgQueue.getFirst();
if (timestamp - head.second >= 10) {
msgQueue.removeFirst();
msgSet.remove(head.first);
} else
break;
}
if (!msgSet.contains(message)) {
Pair<String, Integer> newEntry = new Pair<String, Integer>(message, timestamp);
msgQueue.addLast(newEntry);
msgSet.add(message);
return true;
} else
return false;
}
}
/**
* Your Logger object will be instantiated and called as such:
* Logger obj = new Logger();
* boolean param_1 = obj.shouldPrintMessage(timestamp,message);
*/