刷题146. LRU Cache

一、题目说明

题目146. LRU Cache，设计并实现一个LRU Cache，支持get和put操作。难度是Medium！时间复杂度要求是O(1)。

二、我的解答

时间复杂度要求是O(1)，只能通过hash实现。同时要维护一个容量capacity，当capacity满的时候，更新“最近最少使用的元素”。故需要Hash+LinkedList实现“哈希链表”。

class LRUCache {
	public:
		struct Node{
			int key,value;
			Node* next,*pre;
		};
		Node* head,*rear;

		LRUCache(int size){
			capacity = size;
			head = new Node();
	        rear = new Node();
	        head->pre = NULL;
	        head->next = rear;
	        rear->next = NULL;
	        rear->pre = head;
		}
		int get(int key){
			if(cache.find(key)==cache.end()) return -1;
			Node* tmp = cache[key];

			//移除该节点 
	        tmp->pre->next = tmp->next;
	        tmp->next->pre = tmp->pre;
			
			//插入链头 
	        head->next->pre = tmp;
	        tmp->next = head->next;
	        tmp->pre = head;
	        head->next = tmp;
	        return tmp->value;
		}
		void lru_delete() {
	        if(cache.size() == 0) return;
	        Node* tmp = rear->pre;
	        rear->pre = tmp->pre;
	        tmp->pre->next = rear;
	        cache.erase(tmp->key);
	        delete tmp;
   		}
		void put(int key,int value){
			if(cache.find(key) != cache.end()) {
				//key已存在于链表中，更新值 
	            cache[key]->value = value;
	            this->get(key);
	            return;
        	}
			
			//插入链表中 
	        if(cache.size() >= capacity) 
				this->lru_delete();
	        Node *tmp = new Node;
	        tmp->key = key;
	        tmp->value = value;
	        tmp->pre = this->head;
	        tmp->next = this->head->next;
	        if(head->next != NULL) head->next->pre = tmp;
	        this->head->next = tmp;
	        cache.insert(pair<int, Node*>(key, tmp));
		}
	private:
		map<int,Node*> cache;
		//最大容量 
		int capacity;
};

Runtime: 120 ms, faster than 46.13% of C++ online submissions for LRU Cache.
Memory Usage: 38.1 MB, less than 74.39% of C++ online submissions for LRU Cache.

三、优化措施

无