146. LRU Cache

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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?

Example:

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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

解法1： HashMap + Deque

O(1)的put和get，hashmap应该是逃不了的。问题是还要维护每一个value的access的时间的先后顺序，一般的解法给出来的办法是用一个deque，最近access过的东西放在deque的最后面，而head就存放将要被删除的value。
那么自己要实现的deque的操作就是move_to_head还有一个是remove_from_head，写的时候别忘了当capacity满了之后，需要从deque和hashmap两个地方都删除掉才行。

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class LRUCache {
    class Node {
        int key;
        int val;
        Node prev;
        Node next;
        public Node(int key, int val) {
            this.key = key;
            this.val = val;
            this.prev = null;
            this.next = null;
        }
    };
    private int capacity;
    private HashMap<Integer, Node> map = new HashMap<>();
    private Node head = new Node(-1, -1);
    private Node tail = new Node(-1, -1);
    public LRUCache(int capacity) {
        this.capacity = capacity;
        head.next = tail;
        tail.prev = head;
    }
    public int get(int key) {
        if (!map.containsKey(key)) {
            return -1;
        }
        Node node = map.get(key);
        // Remove from the list
        node.prev.next = node.next;
        node.next.prev = node.prev;
        // move to tail
        move_to_tail(node);
        return node.val;
    }
    public void put(int key, int value) {
        if (get(key) != -1) {
            map.get(key).val = value;
            return;
        }
        if (map.size() == capacity) {
            map.remove(head.next.key);
            remove_from_head();
        }
        Node target = new Node(key, value);
        map.put(key, target);
        move_to_tail(target);
    }
    private void move_to_tail(Node node) {
        node.prev = tail.prev;
        tail.prev = node;
        node.prev.next = node;
        node.next = tail;
    }
    private void remove_from_head() {
        head.next = head.next.next;
        head.next.prev = head;
    }
}
/**
 * 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);
 */