Open icodeish opened 2 years ago
var LRUCache = function (capacity) {
this.capacity = capacity
this.map = new Map()
}
LRUCache.prototype.get = function (key) {
if (this.map.has(key)) {
let value = this.map.get(key)
this.map.delete(key)
this.map.set(key, value)
return value
} else {
return -1
}
}
LRUCache.prototype.put = function (key, value) {
if (this.map.has(key)) {
this.map.delete(key)
}
this.map.set(key, value)
if (this.map.size > this.capacity) {
this.map.delete(this.map.keys().next().value)
}
}
class LRUCache {
capacity: number
map: Map<number, number>
constructor(capacity: number) {
this.capacity = capacity;
this.map = new Map();
}
get(key: number): number {
if (this.map.has(key)) {
const val = this.map.get(key)
this.map.delete(key)
this.map.set(key, val)
return val;
}
return -1;
}
put(key: number, value: number): void {
if(this.map.has(key)) {
this.map.delete(key);
}
this.map.set(key, value);
if (this.map.size > this.capacity) {
this.map.delete([...this.map.keys()][0])
}
}
}
/**
* 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)
*/
class LRUCache {
capacity:number
map: Map<number, number>
constructor(capacity: number) {
this.capacity = capacity
this.map = new Map()
}
get(key: number): number {
if(this.map.has(key)){
const val = this.map.get(key)
this.map.delete(key)
this.map.set(key,val)
return val
}
return -1
}
put(key: number, value: number): void {
if(this.map.has(key)){
this.map.delete(key)
}
this.map.set(key,value)
if(this.capacity < this.map.size){
// map.keys()迭代器
this.map.delete([...this.map.keys()][0])
}
}
}
https://leetcode-cn.com/problems/lru-cache/
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]