Design and implement a data structure for Least Recently Used (LRU) cache. It should support the following operations:getandput.

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 inO(1)time complexity?

Example:

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

class LRUCache {
    class Node{
        Node pre;
        Node next;
        int key;
        int val;

        public Node(int key, int val) {
            this.key = key;
            this.val = val;
        }
    }

    int capacity;
    Map<Integer, Node> map;
    Node head;
    Node tail;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        this.map = new HashMap<>();
        this.head = new Node(-1, -1);
        this.tail = new Node(-1, -1);
        head.next = tail;
        tail.pre = head;
    }

    public int get(int key) {
        if(!map.containsKey(key)) return -1;
        Node cur = map.get(key);

        cur.pre.next = cur.next;
        cur.next.pre = cur.pre;

        tail.pre.next = cur;
        cur.pre = tail.pre;
        tail.pre = cur;
        cur.next = tail;

        return map.get(key).val;
    }

    public void put(int key, int value) {
        if (get(key) != -1) { //map里已经存在这个key，那就把value更新一下就好
                              //且这里get操作了一次，下面在这个if里不用再移动
            Node cur = map.get(key);
            cur.val = value;
            return;
        }

        if(map.size() >= capacity) { //map size大于capacity时候，要在node链表里移除最头上的node，同时移除map里面对应的那个key
            map.remove(head.next.key);
            head.next = head.next.next;
            head.next.pre = head;
        }
        Node node = new Node(key, value); //node里面存的key的值和map里面key值相同   
        map.put(key, node);

        tail.pre.next = node;
        node.pre = tail.pre;
        tail.pre = node;
        node.next = tail;

    }
}