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(…

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…

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.…

题目大意:设计一个用于LRU cache算法的数据结构. 题目链接.关于LRU的基本知识可参考here 分析:为了保持cache的性能,使查找,插入,删除都有较高的性能,我们使用双向链表(std::list)和哈希表(std::unordered_map)作为cache的数据结构,因为: 双向链表插入删除效率高(单向链表插入和删除时,还要查找节点的前节点) 哈希表保存每个节点的地址,可以基本保证在O(1)时间内查找节点 具体实现细节: 越靠近链表头部,表示节点上次访问距离现在时间最短,尾部的节点…

Description: 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 r…

[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…

Problem 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…

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 ret…

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…

class Node(object): def __init__(self,k,x): self.key=k self.val=x self.prev=None self.next=None class DoubleLinkedList(object): def __init__(self): self.tail=None self.head=None def isEmpty(self): return not self.None def removeLast(self): self.remov…