设计实现一个LRU Cache1 什么是LRU Cache2 实现思路

/**
 LRU Cache
 题目描述：
 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?
  
 思路：
 双向链表和hashmap。
 1.当需要插入新的数据项的时候，如果新数据项在链表中存在（一般称为命中），则把该节点移到链表头部，
 如果不存在，则新建一个节点，放到链表头部
 若缓存满了，则把链表最后一个节点删除即可。
 2.在访问数据的时候，如果数据项在链表中存在，则把该节点移到链表头部，否则返回-1。
 这样一来在链表尾部的节点就是最近最久未访问的数据项。
  
 */
public class LRUCache<K , V> {

    class Node<K,V> {
        Node pre;
        Node next;
        private final K key;
        V val;

        Node(K k, V v) {
            key = k;
            val = v;
        }
    }

    Map<K, Node> map = new HashMap<K, Node>();

    // The head (eldest) of the doubly linked list.
    Node head;
    // The tail (youngest) of the doubly linked list.
    Node tail;

    int cap;

    public LRUCache(int capacity) {
        cap = capacity;
        head = new Node(null, null);
        tail = new Node(null, null);
        head.next = tail;
        tail.pre = head;
    }

    public V get(K key) {
        Node n = map.get(key);
        if(n!=null) {
            removeNode(n);
            appendTail(n);
            return (V) n.val;
        }
        return null;
    }

    public void set(K key, V value) {
        Node n = map.get(key);
        // existed
        if(n!=null) {
            n.val = value;
            map.put(key, n);
            removeNode(n);
            appendTail(n);
            return;
        }

        if(map.size() == cap) {
           removeLast();
        }
        n = new Node(key, value);
        // youngest node append tail
        appendTail(n);
        map.put(key, n);
    }

    //移除最近最少使用的节点，注意此节点就是head.next指向的节点
    public void removeLast() {
        Node tmp = head.next;
        removeNode(tmp);
        map.remove(tmp.key);
    }

    //移除某个节点，并将此节点的前后节点连接起来
    private void removeNode(Node n){
        n.pre.next = n.next;
        n.next.pre = n.pre;
    }

    //将节点加入到队列的尾部，尾部代表最近使用的节点
    private void appendTail(Node n) {
        n.next = tail;
        n.pre = tail.pre;
        tail.pre.next = n;
        tail.pre = n;
    }
}

public class LRUCache<K , V> {

    private int capacity;
    private Map<K, V> cache;

    public LRUCache(final int capacity) {
        this.capacity = capacity;
        this.cache = new java.util.LinkedHashMap<K, V> (capacity, 0.75f, true) {
            // 定义put后的移除规则，大于容量就删除eldest
            protected boolean removeEldestEntry(Map.Entry<K, V> eldest) {
                return size() > capacity;
            }
        };
    }

    public V get(K key) {
        if (cache.containsKey(key)) {
            return cache.get(key);
        } else
            return null;
    }

    public void set(K key, V value) {
        cache.put(key, value);
    }
}