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…

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…

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…

设计并实现最不经常使用(LFU)缓存的数据结构.它应该支持以下操作:get 和 put. get(key) - 如果键存在于缓存中,则获取键的值(总是正数),否则返回 -1.put(key, value) - 如果键不存在,请设置或插入值.当缓存达到其容量时,它应该在插入新项目之前,使最不经常使用的项目无效.在此问题中,当存在平局(即两个或更多个键具有相同使用频率)时,最近最少使用的键将被去除. 进阶:你是否可以在 O(1) 时间复杂度内执行两项操作? 示例: LFUCache cache =…

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 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(…

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…

设计并实现最近最久未使用(Least Recently Used)缓存. 题目描述: 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 ex…

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…

https://leetcode.com/problems/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 exis…

题目:LRU cache Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations: get and set. . set(key, value) - Set or insert the value if the key is not already present. When the cache reached its…

插话:只写了几个连续的博客,博客排名不再是实际"远在千里之外"该.我们已经进入2一万内. 再接再厉.油! 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 ke…

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(…

题目简述: 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 -…

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…

esign 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(k…

原题地址 以前Leetcode的测试数据比较弱,单纯用链表做也能过,现在就不行了,大数据会超时.通常大家都是用map+双向链表做的. 我曾经尝试用C++的list容器来写,后来发现map没法保存list的iterator,总是报错,我也不知道为什么.后来只好手写双向链表,真是痛苦啊,一不小心就会出错.怪不得这道题是hard 代码: struct Node { int key; int val; Node *prev; Node *next; Node() : prev(NULL), next(N…

很实际的一道题.定义一个双向链表list,方便插入和删除:定义一个哈希表,方便查找. 具体的,哈希表存放每个结点的key和它对应的结点的地址:访问结点时,如果结点存在,则将其交换到头部,同是更新哈希表中的地址: 插入结点时,首先判断capacity是否达到了上限,如果是则在链表和哈希表中删除该结点:新结点插入链表头部. 有很多细节需要注意,双向链表和哈希表的用法,需要多加体会. class LRUCache { private: struct CacheNode { int key; int v…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 477 Total Hamming Distance 44.10% Meidum 475 Heaters  30.20% Easy 474 Ones and Zeroes  34.90% Meidum 473 Matchsticks to Square  31.80% Medium 472 Concatenated Words 29.20% Hard…

82. Remove Duplicates from Sorted List II https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/ 下面代码是自写hashmap版,练一下hashmap,用一般的map<int,int>红黑树map也能过. /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *…

突然很想刷刷题,LeetCode是一个不错的选择,忽略了输入输出,更好的突出了算法,省去了不少时间. dalao们发现了任何错误,或是代码无法通过,或是有更好的解法,或是有任何疑问和建议的话,可以在对应的随笔下面评论区留言,我会及时处理,在此谢过了. 过程或许会很漫长,也很痛苦,慢慢来吧. 编号 题名 过题率 难度 1 Two Sum 0.376 Easy 2 Add Two Numbers 0.285 Medium 3 Longest Substring Without Repeating C…

All LeetCode Questions List(Part of Answers, still updating) 题目汇总及部分答案(持续更新中) Leetcode problems classified by company 题目按公司分类(Last updated: October 2, 2017) .   Top Interview Questions # Title Difficulty Acceptance 1 Two Sum Medium 17.70% 2 Add Two N…

All LeetCode Questions List 题目汇总 Sorted by frequency of problems that appear in real interviews. Last updated: October 2, 2017Google (214)534 Design TinyURL388 Longest Absolute File Path683 K Empty Slots340 Longest Substring with At Most K Distinct C…

源代码地址:https://github.com/hopebo/hopelee 语言:C++ 301. Remove Invalid Parentheses Remove the minimum number of invalid parentheses in order to make the input string valid. Return all possible results. Note: The input string may contain letters other tha…

终于将LeetCode的免费题刷完了,真是漫长的第一遍啊,估计很多题都忘的差不多了,这次开个题目汇总贴,并附上每道题目的解题连接,方便之后查阅吧~ 如果各位看官们,大神们发现了任何错误,或是代码无法通过OJ,或是有更好的解法,或是有任何疑问,意见和建议的话,请一定要在对应的帖子下面评论区留言告知博主啊(如果不方便注册博客园的话,可以下载下文提到的APP,在Feedback中给博主发邮件交流哈),同时也请大家踊跃地,大量地,盲目地提供各个题目的follow up一起讨论哈,多谢多谢,祝大家刷得愉快…

43. Multiply Strings 高精度非负整数的乘法. string multiply(string num1, string num2) { '); ; <= i; --i) { ; ; <= j; --j) { ] - ') + carry; sum[i + j + ] = tmp % + '; carry = tmp / ; } sum[i] += carry; } size_t startpos = sum.find_first_not_of("); if (str…