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 <= 1 0 5 10^5 105
- 最多调用 2 * 1 0 5 10^5 105 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: number
constructor(capacity: number) {
this.cache = new Map()
this.max = capacity
}
get(key: number): number {
if (!this.cache.has(key)) return -1
const 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)
*/
