Design and implement a data structure for Least Frequently Used (LFU) 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.pu…

Design and implement a data structure for Least Frequently Used (LFU) 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. s…

原题链接在这里:https://leetcode.com/problems/lfu-cache/?tab=Description 题目: Design and implement a data structure for Least Frequently Used (LFU) cache. It should support the following operations: getand put. get(key) - Get the value (will always be positiv…

Design and implement a data structure for Least Frequently Used (LFU) 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.pu…

LRU算法是首先淘汰最长时间未被使用的页面,而LFU是先淘汰一定时间内被访问次数最少的页面,如果存在使用频度相同的多个项目,则移除最近最少使用(Least Recently Used)的项目. LFU在频度相同的时候与LRU类似. 146. LRU Cache 1.stl中list是双向链表,slist才是单向链表. rbegin():返回尾元素的逆向迭代器指针 end():返回尾元素之后位置的迭代器指针 https://www.cnblogs.com/Kobe10/p/5780095.html…

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…

2018-11-06 20:06:04 LFU(Least Frequently Used)算法根据数据的历史访问频率来淘汰数据,其核心思想是“如果数据过去被访问多次,那么将来被访问的频率也更高”. 如何高效的实现一个LFU Cache是一个难点,其实现方式要比LRU要复杂一点,问题的核心就是如果对不同的freq进行计数和维护.这里用到的思路和最大频率栈是类似的,也就是对每个freq都开辟一个Set来进行单独的维护. 为了实现的方便,我们可以在每个freq节点中放入一个LinkedHashSet…

在Leetcode上遇到了两个有趣的题目,分别是利用LRU和LFU算法实现两个缓存.缓存支持和字典一样的get和put操作,且要求两个操作的时间复杂度均为O(1). 首先说一下如何在O(1)时间复杂度内实现get方法.据鄙人所知,对于没有限定数据范围的数据,唯一拥有O(1)时间复杂度的get的数据结构就是哈希表,尽管其时间复杂度是通过概率来推算出来的.因此毋庸质疑,LRU和LFU的get方法中必定使用了哈希表的取值,相应的put方法中也必定调用了哈希表的赋值操作. 由于LRU和LFU都涉及到了选…

hash:存储的key.value.freq freq:存储的freq.key,也就是说出现1次的所有key在一起,用list连接 class LFUCache { public: LFUCache(int capacity) { cap = capacity; } int get(int key) { auto it = hash.find(key); if(it == hash.end()) ; 这段代码以下处理的都是已有key的情况 freq[hash[key].second].erase…

题目: 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.…