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LRU Cache
题意:
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.
学LRU是在操作系统中,最近最久未被使用的,根据cache的大小,每当有一个进程来到时,都要根据当前的情况而做一些改变。
set:如果当前来到的key在cache中,则将该节点删除并且移到链表的头部,如果不存在,并且当前的cache满了,则将尾部删除,将新来的节点加入到链表的首部,否则直接加入练表中即可。
get:如果当前的key在cache中,则将该节点从链表中移除并且添加带链表的首部并返回value,否则返回-1。
代码给的是用hashmap和双链表完成的,因为当删除一个节点的时候,需要知道他的后记节点,虽然单链表也能做到这一点,但是如果要删除的节点是尾节点,那么就要知道他的前一个节点,这样的话就要便利整个链表,那么就会花o(n)的时间复杂度,虽然说当链表中有n个节点,那么(o(1)*(n-1) + o(n))/n这样平均的话时间复杂度是在o(1)的,但是也可以彻底做到时间复杂度是o(1)的,就用双链表,会将首位标记出来。
struct Linklist { int key; int value; struct Linklist* pre; struct Linklist* next; }; class LRUCache { private: int count; int size; struct Linklist* tail; struct Linklist* head; map<int,Linklist*>mp; public: LRUCache(int size) { this->size = size; this->tail = NULL; this->head = NULL; this->count = 0; } //如果该节点在map中存在,则删除当前的,并改变节点的值,将该节点放在双链表的表头 //否则返回-1 int get(int key) { if(head == NULL) return -1; map<int,Linklist*>::iterator it = mp.find(key); if(it == mp.end()) { return -1; } else { Linklist *p = it->second; pushup(p); } return head->value; } //双链表删除节点 然后放在双链表的表头 void pushup(Linklist* p) { if(this->count == 1) return ; if(p == this->head) return; if(p == this->tail) { tail = p->pre; } p->pre->next = p->next; if(p->next != NULL) p->next->pre = p->pre; p->next = head; p->pre = NULL; head->pre = p; head = p; } //当前值,没有在cache中 则删除链表中最后一个节点&&map中删除,然后将访问的节点加在链表的表头 //没有在cache中,则将尾删除 将新的元素入链表 并将元素加入到map中 void set(int key,int value) { if(head == NULL) { head = (Linklist *)malloc(sizeof(Linklist)); head->value =http://www.mamicode.com/ value; head->key = key; head->pre = NULL; head->next = NULL; mp[key] = head; tail = head; this->count++; } else { map<int,Linklist*>::iterator it = mp.find(key); if(it == mp.end()) { if(size == count) { if(size == 1 && tail == head) //size == 1 { mp.erase(head->key); head->key = key; head->value =http://www.mamicode.com/ value; mp[head->key] = head; } else //满 { Linklist *p = tail; tail->pre->next = p->next; mp.erase(tail->key); tail = tail->pre; p->key = key; p->value =http://www.mamicode.com/ value; p->next = head; head->pre = p; head = p; mp[head->key] = head; } } else { Linklist *p = (Linklist*)malloc(sizeof(Linklist)); p->value =http://www.mamicode.com/ value; p->key = key; p->next = head; p->pre = NULL; head->pre = p; head = p; mp[head->key] = head; this->count++; } } else { Linklist *p = it->second; p->value =http://www.mamicode.com/ value; pushup(p); } } } };
lala
LRU Cache