【LeetCode】38. LRU Cache

题目描述(Hard)

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(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.

Follow up:
Could you do both operations in O(1) time complexity?

题目链接

https://leetcode.com/problems/lru-cache/description/

Example 1:

LRUCache cache = new LRUCache( 2 /* capacity */ );

cache.put(1, 1);
cache.put(2, 2);
cache.get(1);       // returns 1
cache.put(3, 3);    // evicts key 2
cache.get(2);       // returns -1 (not found)
cache.put(4, 4);    // evicts key 1
cache.get(1);       // returns -1 (not found)
cache.get(3);       // returns 3
cache.get(4);       // returns 4

算法分析

使用双向链表及哈希表保证查找、插入及删除都有较高的性能：

• 哈希表保存每个节点地址，可以基本保证在O(1)时间内查找节点；
• 双向链表插入和删除的效率高，单向链表插入和删除时，还要查找节点的前驱节点；

具体实现方法：

• 越靠近链表的头部，表示节点上次访问距离现在时间最短，尾部的节点表示最近访问最少；
• 访问节点时，如果节点存在，把该节点交换到链表头部，同时更新hash表中的该节点的地址；
• 插入节点时，如果cache的size达到了上限capacity，则删除尾部节点，同时要在hash表中删除对应的项，新节点插入链表头部。

提交代码：

class LRUCache {
public:
	LRUCache(int capacity) {
		this->capacity = capacity;
	}

	int get(int key) {
		if (cacheMap.find(key) == cacheMap.end()) return -1;
		cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);
		cacheMap[key] = cacheList.begin();
		return cacheMap[key]->value;
	}

	void put(int key, int value) {
		if (cacheMap.find(key) == cacheMap.end())
		{
			// 满容量则删除尾部元素
			if (cacheMap.size() == capacity)
			{
				cacheMap.erase(cacheList.back().key);
				cacheList.pop_back();
			}
			// 在头部插入元素
			cacheList.push_front(CacheNode(key, value));
			cacheMap[key] = cacheList.begin();
		}
		else
		{
			cacheMap[key]->value = value;
			cacheList.splice(cacheList.begin(), cacheList, cacheMap[key]);
			cacheMap[key] = cacheList.begin();
		}
	}

private:
	struct CacheNode {
		int key;
		int value;
		CacheNode(int k, int v) : key(k), value(v) {}
	};
	list<CacheNode> cacheList;
	unordered_map<int, list<CacheNode>::iterator> cacheMap;
	int capacity;
};

/**
 * Your LRUCache object will be instantiated and called as such:
 * LRUCache obj = new LRUCache(capacity);
 * int param_1 = obj.get(key);
 * obj.put(key,value);
 */