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0146. LRU Cache

难度: Medium

刷题内容

原题链接

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

内容描述

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

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1. put(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.

The cache is initialized with a positive capacity.

Follow up: Could you do both operations in O(1) time complexity?

Example:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

解题方案

思路1

**- 时间复杂度: O(1)*- 空间复杂度: O(N)***

这是一道数据结构的考题。

简单的获取和插入，顺序不做考虑，所以数据结构用Object即可。

但是重要的考点——LRU，也就是删除最近没有使用的数据。所以想到用队列存在key列表：

  • 未超过存储上限时，每次put一个新数据时，向队列末尾插入当前key
  • 每次get时，如果key存在，则将对应的key从的队列中，移到对末尾
  • 在超过存储上限时，如进行put操作，则将队列首位删除掉

执行用时 :492 ms, 在所有 JavaScript 提交中击败了35.29%的用户

内存消耗 :59.7 MB, 在所有 JavaScript 提交中击败了16.36%的用户

/**
 * @param {number} capacity
 */
var LRUCache = function(capacity) {
  this.limit = capacity || 2
  this.storage = {}
  this.keyList = []
};

/**
 * @param {number} key
 * @return {number}
 */
LRUCache.prototype.get = function(key) {
  if (this.storage.hasOwnProperty(key)) {
    let index = this.keyList.findIndex(k => k === key)
    this.keyList.splice(index, 1)
    this.keyList.push(key)
    return this.storage[key]
  } else {
    return -1
  }
};

/**
 * @param {number} key
 * @param {number} value
 * @return {void}
 */
LRUCache.prototype.put = function(key, value) {
  // 判断容量
  if (this.keyList.length >= this.limit && !this.storage.hasOwnProperty(key)) {
    this.deleteLRU()
  }

  // 存储数据
  this.updateKeyList(key)
  this.storage[key] = value
};

LRUCache.prototype.deleteLRU = function () {
  delete this.storage[this.keyList.shift()]
}

LRUCache.prototype.updateKeyList = function (key) {
  if (this.storage.hasOwnProperty(key)) {
    var index = this.keyList.findIndex(k => key === k)
    this.keyList.splice(index, 1)
  }
  this.keyList.push(key)
}

/**
 * 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)
 */
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