LRU Cache

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

get(key) - Get the value (will always be positive) of the key if the key exists in the cache, otherwise return -1.

set(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.

Tips:

考察对基本数据结构哈希表、链表的掌握，

为了使查找、插入和删除都有较高的性能，首先定义一个Node节点存储key, value, prev, next。使用一个哈希表。每次进行操作以后把最近操作的放到队尾。

复杂度：

get（）采用哈希表查找，时间复杂度O（1），set（）采用双向链表，删除插入时间复杂度O(1)

Code:

public class LRUCache {
    private class Node{
        Node prev;
        Node next;
        int key;
        int value;

        public Node(int key, int value) {
            this.key = key;
            this.value = value;
            this.prev = null;
            this.next = null;
        }
    }

    private int capacity;
    private HashMap<Integer, Node> hs = new HashMap<Integer, Node>();
    private Node head = new Node(-1, -1);
    private Node tail = new Node(-1, -1);

    public LRUCache(int capacity) {
        this.capacity = capacity;
        tail.prev = head;
        head.next = tail;
    }

    public int get(int key) {
        if( !hs.containsKey(key)) {
            return -1;
        }

        // remove current
        Node current = hs.get(key);
        current.prev.next = current.next;
        current.next.prev = current.prev;

        // move current to tail
        move_to_tail(current);

        return hs.get(key).value;
    }

    public void set(int key, int value) {
        if( get(key) != -1) {
            hs.get(key).value = value;
            return;
        }

        if (hs.size() == capacity) {
            hs.remove(head.next.key);
            head.next = head.next.next;
            head.next.prev = head;
        }

        Node insert = new Node(key, value);
        hs.put(key, insert);
        move_to_tail(insert);
    }

    private void move_to_tail(Node current) {
        current.prev = tail.prev;
        tail.prev = current;
        current.prev.next = current;
        current.next = tail;
    }
}