Leetcode(146) LRU Cache

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Description

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.

The cache is initialized with a positive capacity.

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

用一个哈希表和一个双向链表实现，注意删除节点，移动节点时头尾节点的处理，同时如果put一个已有的key对，则更新并置于队尾即可。写的真心心累…

具体代码如下：

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class LRUCache {

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

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

    Node head = null;
    Node tail = null;
    int capacity;
    HashMap<Integer, Node> valMap;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        this.valMap = new HashMap<>();
    }

    public int get(int key) {
        if (this.valMap.containsKey(key)) {
            Node result = this.valMap.get(key);
            Node pre = result.pre;
            Node next = result.next;
            if (head == result && tail == result) {
                return result.val;
            }
            if (head == result) {
                head = next;
                tail.next = result;
                result.pre = tail;
                result.next = null;
                tail = result;
                return result.val;
            } else if (tail == result) {
                return result.val;
            } else {
                pre.next = next;
                next.pre = pre;
                tail.next = result;
                result.pre = tail;
                result.next = null;
                tail = result;
                return result.val;
            }
        }
        return -1;
    }

    public void put(int key, int value) {
        if (valMap.containsKey(key)) {
            this.get(key);
            tail.val = value;
            return;
        }
        Node node = new Node(key, value);
        if (this.valMap.size() < this.capacity) {
            if (head == null) {
                head = node;
                head.pre = null;
                head.next = null;
                tail = node;
            } else {
                tail.next = node;
                node.pre = tail;
                node.next = null;
                tail = node;
            }
        } else {
            valMap.remove(head.key);
            if (head == tail) {
                head = tail = node;
                this.valMap.put(key, node);
                return;
            }
            head = head.next;
            head.pre = null;
            tail.next = node;
            node.pre = this.tail;
            node.next = null;
            tail = node;

        }
        this.valMap.put(key, node);
    }

}

/**
 * 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);
 */