2622. 有时间限制的缓存

🟠   🔗 力扣 LeetCode

题目

Write a class that allows getting and setting key-value pairs, however a time until expiration is associated with each key.

The class has three public methods:

set(key, value, duration): accepts an integer key, an integer value, and a duration in milliseconds. Once the duration has elapsed, the key should be inaccessible. The method should return true if the same un-expired key already exists and false otherwise. Both the value and duration should be overwritten if the key already exists.

get(key): if an un-expired key exists, it should return the associated value. Otherwise it should return -1.

count(): returns the count of un-expired keys.

Example 1:

Input:

actions = ["TimeLimitedCache", "set", "get", "count", "get"]

values = [[], [1, 42, 100], [1], [], [1]]

timeDelays = [0, 0, 50, 50, 150]

Output: [null, false, 42, 1, -1]

Explanation:

At t=0, the cache is constructed.

At t=0, a key-value pair (1: 42) is added with a time limit of 100ms. The value doesn't exist so false is returned.

At t=50, key=1 is requested and the value of 42 is returned.

At t=50, count() is called and there is one active key in the cache.

At t=100, key=1 expires.

At t=150, get(1) is called but -1 is returned because the cache is empty.

Example 2:

Input:

actions = ["TimeLimitedCache", "set", "set", "get", "get", "get", "count"]

values = [[], [1, 42, 50], [1, 50, 100], [1], [1], [1], []]

timeDelays = [0, 0, 40, 50, 120, 200, 250]

Output: [null, false, true, 50, 50, -1, 0]

Explanation:

At t=0, the cache is constructed.

At t=0, a key-value pair (1: 42) is added with a time limit of 50ms. The value doesn't exist so false is returned.

At t=40, a key-value pair (1: 50) is added with a time limit of 100ms. A non-expired value already existed so true is returned and the old value was overwritten.

At t=50, get(1) is called which returned 50.

At t=120, get(1) is called which returned 50.

At t=140, key=1 expires.

At t=200, get(1) is called but the cache is empty so -1 is returned.

At t=250, count() returns 0 because the cache is empty.

Constraints:

• 0 <= key, value <= 10^9
• 0 <= duration <= 1000
• 1 <= actions.length <= 100
• actions.length === values.length
• actions.length === timeDelays.length
• 0 <= timeDelays[i] <= 1450
• actions[i] is one of "TimeLimitedCache", "set", "get" and "count"
• First action is always "TimeLimitedCache" and must be executed immediately, with a 0-millisecond delay

题目大意

编写一个类，它允许获取和设置键-值对，并且每个键都有一个 过期时间 。

该类有三个公共方法：

set(key, value, duration) ：接收参数为整型键 key 、整型值 value 和以毫秒为单位的持续时间 duration 。一旦 duration 到期后，这个键就无法访问。如果相同的未过期键已经存在，该方法将返回 true ，否则返回 false 。如果该键已经存在，则它的值和持续时间都应该被覆盖。

get(key) ：如果存在一个未过期的键，它应该返回这个键相关的值。否则返回 -1 。

count() ：返回未过期键的总数。

提示：

• 0 <= key, value <= 10^9
• 0 <= duration <= 1000
• 1 <= actions.length <= 100
• actions.length === values.length
• actions.length === timeDelays.length
• 0 <= timeDelays[i] <= 1450
• actions[i] 是 "TimeLimitedCache"、"set"、"get" 和 "count" 中的一个。
• 第一个操作始终是 "TimeLimitedCache" 而且一定会以 0 毫秒的延迟立即执行

解题思路

1. constructor：

   • 使用 Map 数据结构来存储键值对，其中每个键对应的值为一个数组 [value, timer]，其中 value 是存储的值，timer 是该键的定时器，用于删除过期键。

2. set(key, value, duration)：

   • 该方法将给定的键 key 和值 value 存入缓存，并设置该键值对的有效期 duration 毫秒。
   • 如果相同键已经存在且未过期，则返回 true，否则返回 false。
   • 使用 setTimeout 定时函数，在到期时间后自动删除该键值对。
   • 如果键已经存在且定时器没有到期，首先清除之前的定时器，以确保新设置的 duration 覆盖旧的时间。

3. get(key)：

   • 查询缓存中是否存在未过期的键 key。
   • 如果键存在且未过期，则返回其对应的值，否则返回 -1。

4. count()：

   • 返回当前缓存中未过期的键值对数量，直接通过 Map 对象的 size 属性来获取当前键的数量。

复杂度分析

• 时间复杂度：
  • set：每次调用 set，查找和插入操作的时间复杂度为 O(1)，删除旧的定时器和设置新的定时器也为常数时间操作。因此 set 的时间复杂度为 O(1)。
  • get：查找键的时间复杂度为 O(1)。
  • count：Map.size 的访问是常数时间操作，时间复杂度为 O(1)。
• 空间复杂度：O(n)，其中 n 为缓存中存储的键值对数量，使用了 Map 来存储键值对。

代码

var TimeLimitedCache = function () {
	this.cache = new Map();
};

/**
 * @param {number} key
 * @param {number} value
 * @param {number} duration time until expiration in ms
 * @return {boolean} if un-expired key already existed
 */
TimeLimitedCache.prototype.set = function (key, value, duration) {
	const timer = setTimeout(() => this.cache.delete(key), duration);
	const exist = this.cache.has(key);
	if (exist) {
		const oldTimer = this.cache.get(key)[1];
		clearTimeout(oldTimer);
	}
	this.cache.set(key, [value, timer]);
	return exist;
};

/**
 * @param {number} key
 * @return {number} value associated with key
 */
TimeLimitedCache.prototype.get = function (key) {
	if (this.cache.has(key)) return this.cache.get(key)[0];
	return -1;
};

/**
 * @return {number} count of non-expired keys
 */
TimeLimitedCache.prototype.count = function () {
	return this.cache.size;
};

/**
 * const timeLimitedCache = new TimeLimitedCache()
 * timeLimitedCache.set(1, 42, 1000); // false
 * timeLimitedCache.get(1) // 42
 * timeLimitedCache.count() // 1
 */

相关题目

题号	标题	题解	标签	难度	力扣
2627	函数防抖	[✓]		🟠	🀄️ 🔗
2636	Promise 对象池 🔒	[✓]		🟠	🀄️ 🔗
2637	有时间限制的 Promise 对象	[✓]		🟠	🀄️ 🔗