LRU 缓存

• https://leetcode.cn/problems/lru-cache/description/

描述

• 请你设计并实现一个满足 LRU (最近最少使用) 缓存 约束的数据结构
  实现 LRUCache 类：
  • LRUCache(int capacity) 以 正整数 作为容量 capacity 初始化 LRU 缓存
  • int get(int key) 如果关键字 key 存在于缓存中，则返回关键字的值，否则返回 -1
  • void put(int key, int value) 如果关键字 key 已经存在，则变更其数据值 value ；如果不存在，则向缓存中插入该组 key-value
  • 如果插入操作导致关键字数量超过 capacity ，则应该 逐出 最久未使用的关键字
• 函数 get 和 put 必须以 O(1) 的平均时间复杂度运行

示例

输入
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
输出
[null, null, null, 1, null, -1, null, -1, 3, 4]

解释
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // 缓存是 {1=1}
lRUCache.put(2, 2); // 缓存是 {1=1, 2=2}
lRUCache.get(1); // 返回 1
lRUCache.put(3, 3); // 该操作会使得关键字 2 作废，缓存是 {1=1, 3=3}
lRUCache.get(2); // 返回 -1 (未找到)
lRUCache.put(4, 4); // 该操作会使得关键字 1 作废，缓存是 {4=4, 3=3}
lRUCache.get(1); // 返回 -1 (未找到)
lRUCache.get(3); // 返回 3
lRUCache.get(4); // 返回 4

提示

• 1 <= capacity <= 3000
• 0 <= key <= 10000
• 0 <= value <= $10^5$
• 最多调用 2 * $10^5$ 次 get 和 put

Typescript 版算法实现

1 ) 方案1: 基于Map

class LRUCache {public cache: Map<number, any>public max: numberconstructor(capacity: number) {this.cache = new Map()this.max = capacity}get(key: number): number {if (!this.cache.has(key)) return -1const tmp = this.cache.get(key)this.cache.delete(key)this.cache.set(key, tmp)return tmp}put(key: number, value: number): void {if(this.cache.has(key)) {this.cache.delete(key)} else if(this.cache.size >= this.max) {// 新增时的淘汰机制this.cache.delete(this.cache.keys().next().value)}this.cache.set(key, value)}
}/*** Your LRUCache object will be instantiated and called as such:* var obj = new LRUCache(capacity)* var param_1 = obj.get(key)* obj.put(key,value)*/

2 ) 方案2: 哈希表 + 双向链表

class DLinkedNode {key: number;value: number;prev: DLinkedNode | null;next: DLinkedNode | null;constructor(key: number = 0, value: number = 0) {this.key = key;this.value = value;this.prev = null;this.next = null;}
}class LRUCache {private size: number;private capacity: number;private cache: Map<number, DLinkedNode>;private head: DLinkedNode;private tail: DLinkedNode;constructor(capacity: number) {this.size = 0;this.capacity = capacity;this.cache = new Map();this.head = new DLinkedNode();this.tail = new DLinkedNode();this.head.next = this.tail;this.tail.prev = this.head;}private addToHead(node: DLinkedNode): void {node.prev = this.head;node.next = this.head.next;this.head.next!.prev = node;this.head.next = node;}private removeNode(node: DLinkedNode): void {node.prev!.next = node.next;node.next!.prev = node.prev;}private moveToHead(node: DLinkedNode): void {this.removeNode(node);this.addToHead(node);}private removeTail(): DLinkedNode {const node = this.tail.prev!;this.removeNode(node);return node;}public get(key: number): number {if (!this.cache.has(key)) {return -1;}const node = this.cache.get(key)!;this.moveToHead(node);return node.value;}public put(key: number, value: number): void {if (!this.cache.has(key)) {const newNode = new DLinkedNode(key, value);this.cache.set(key, newNode);this.addToHead(newNode);this.size++;if (this.size > this.capacity) {const removedNode = this.removeTail();this.cache.delete(removedNode.key);this.size--;}} else {const node = this.cache.get(key)!;node.value = value;this.moveToHead(node);}}
}/*** Your LRUCache object will be instantiated and called as such:* var obj = new LRUCache(capacity)* var param_1 = obj.get(key)* obj.put(key,value)*/