[LeetCode][JavaScript]LRU Cache

LRU Cache

Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set.

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.
set(key, value) - Set or insert the value if the key is not already present. When the cache reached its capacity, it should invalidate the least recently used item before inserting a new item.

https://leetcode.com/problems/lru-cache/

 

 

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最近最少使用。

开一个容量为capacity的容器，如果最近访问过(get和set都算访问)，就把这个元素放到第一个，超出capacity就删掉最后一个元素。

我这边用了js的数组，可以AC，但比较慢，用链表可以快一些，主要是移除数组里指定元素的操作费时：

 this.queue = this.queue.slice(0,i).concat(this.queue.slice(i + 1));

另开一个set对象记录在队列中的元素，以空间换时间，这样查找元素的效率就是O(logn)

/**
 * @constructor
 */
var LRUCache = function(capacity) {
    this.capacity = capacity;
    this.keyMap = new Set();
    this.queue = []; 
};

/**
 * @param {number} key
 * @returns {number}
 */
LRUCache.prototype.get = function(key) {
    if(this.keyMap.has(key)){
        for(var i = 0; i < this.queue.length; i++){
            if(this.queue[i].key === key){
                var value = this.queue[i].value;
                this.queue = this.queue.slice(0,i).concat(this.queue.slice(i + 1));
                this.queue.unshift({key : key, value : value}); 
                return value;
            }
        }
    }
    return -1;
};

/**
 * @param {number} key
 * @param {number} value
 * @returns {void}
 */
LRUCache.prototype.set = function(key, value) {
    if(this.keyMap.has(key)){
        var index = -1;
        for(var i = 0; i < this.queue.length; i++){
            if(this.queue[i].key === key){
                index = i;
                break;
            }
        }
        this.queue = this.queue.slice(0,index).concat(this.queue.slice(index + 1));
    }else{
        this.keyMap.add(key);
    }
    this.queue.unshift({key : key, value : value}); 

    if(this.queue.length > this.capacity){
        var pop = this.queue.pop();
        this.keyMap.delete(pop.key);
    }
};