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(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.
模拟近期最少使用算法。首先分析近期最少使用算法的特点:1.需要有一个时间链表,来存储并排序最近使用过的一些单元;2.对于新的单元,要能够查询时间链表中是否存在这个单元,如果存在,则操作这个单元并把这个单元放在时间链表的最前面,如果不存在,则插入这个单元到时间链表的最前面,此时,如果列表已经满了,需要删除队尾元素。
所以,数据结构需要能实现一个时间链表来表示单元的先后顺序,同时需要能尽快的查询到这个链表中是否存在要操作的单元。
查询操作,用map<key,value>结构,时间复杂度是O(1),对于时间链表,我们可以通过构造链表来完成,于是map<key,value>结构应该为map<key,Node>,其中Node为结构体,存储value的同时实现一个链表结构。
代码如下:
class LRUCache{ private: int maxCapacity; int curCapacity; int headKey; int tailKey; struct LRU{ int value; int preKey; int nextKey; LRU(int value):value(value),preKey(-1),nextKey(-1){} }; unordered_map<int, LRU*> lruMap; public: LRUCache(int capacity) { maxCapacity=capacity; curCapacity=0; } int get(int key) { if(lruMap.find(key)==lruMap.end())return -1; else{ if(maxCapacity==1)return lruMap[key]->value; else{ if(headKey==key)return lruMap[key]->value; else if(tailKey==key){ tailKey=lruMap[key]->preKey; lruMap[key]->nextKey=headKey; lruMap[headKey]->preKey=key; headKey=key; return lruMap[key]->value; }else{ int preKey=lruMap[key]->preKey; int nextKey=lruMap[key]->nextKey; lruMap[preKey]->nextKey=nextKey; lruMap[nextKey]->preKey=preKey; lruMap[key]->nextKey=headKey; lruMap[headKey]->preKey=key; headKey=key; return lruMap[key]->value; } } } } void set(int key, int value) { if(maxCapacity<=0)return; if(lruMap.find(key)==lruMap.end()){ LRU* Node = new LRU(value); if(maxCapacity==1){ lruMap.clear(); lruMap.insert(make_pair(key,Node)); }else{ if(curCapacity==0){ lruMap.insert(make_pair(key,Node)); curCapacity++; headKey=tailKey=key; }else if(curCapacity<maxCapacity){ lruMap.insert(make_pair(key,Node)); lruMap[key]->nextKey=headKey; lruMap[headKey]->preKey=key; headKey=key; curCapacity++; }else{ tailKey=lruMap[tailKey]->preKey; lruMap.erase(lruMap[tailKey]->nextKey); lruMap.insert(make_pair(key,Node)); lruMap[key]->nextKey=headKey; lruMap[headKey]->preKey=key; headKey=key; } } }else{ if(maxCapacity==1)lruMap[key]->value=value; else{ if(headKey==key)lruMap[key]->value=value; else if(tailKey==key){ tailKey=lruMap[key]->preKey; lruMap[key]->nextKey=headKey; lruMap[headKey]->preKey=key; headKey=key; lruMap[key]->value=value; }else{ int preKey=lruMap[key]->preKey; int nextKey=lruMap[key]->nextKey; lruMap[preKey]->nextKey=nextKey; lruMap[nextKey]->preKey=preKey; lruMap[key]->nextKey=headKey; lruMap[headKey]->preKey=key; headKey=key; lruMap[key]->value=value; } } } } };