Lintcode24 LFU Cache solution 题解

【题目描述】

LFU (Least Frequently Used) is a famous cache eviction algorithm.For a cache with capacity k, if the cache is full and need to evict a key in it, the key with the lease frequently used will be kicked out.Implement set and get method for LFU cache.

LFU是一个著名的缓存算法。实现LFU中的set 和 get

【题目链接】

http://www.lintcode.com/en/problem/lfu-cache/

【题目解析】

这道题让我们实现一个LRU缓存器，LRU是Least Recently Used的简写，就是最近最少使用的意思。那么这个缓存器主要有两个成员函数，get和set，其中get函数是通过输入key来获得value，如果成功获得后，这对(key, value)升至缓存器中最常用的位置（顶部），如果key不存在，则返回-1。而set函数是插入一对新的(key, value)，如果原缓存器中有该key，则需要先删除掉原有的，将新的插入到缓存器的顶部。如果不存在，则直接插入到顶部。若加入新的值后缓存器超过了容量，则需要删掉一个最不常用的值，也就是底部的值。具体实现时我们需要三个私有变量，cap, l和m，其中cap是缓存器的容量大小，l是保存缓存器内容的列表，m是哈希表，保存关键值key和缓存器各项的迭代器之间映射，方便我们以O(1)的时间内找到目标项。

【参考答案】

http://www.jiuzhang.com/solutions/lfu-cache/

时间： 2024-10-23 21:38:10

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