[146 LRU缓存](https://leetcode.cn/problems/lru-cache/)

分析

维护一个双向链表保存缓存中的元素。

如果元素超过容量阈值，则删除最久未使用的元素。为了实现这个功能，将get(), put()方法获取的元素添加到链表首部。

为了在O(1)时间复杂度执行get()方法，再新建一个映射表，缓存key与链表节点。

源码

class LRUCache {
    private int size;
    private int capacity;

    private Map<Integer, Node> cache = new HashMap<>();
    private Node head;
    private Node tail;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        this.size = 0;
        this.head = new Node();
        this.tail = new Node();
        this.head.next = this.tail;
        this.tail.prev = this.head;
    }
    
    public int get(int key) {
        Node node = cache.get(key);
        if (node == null) {
            return -1;
        }
        moveToHead(node);
        return node.value;
    }
    
    public void put(int key, int value) {
        Node node = cache.get(key);
        if (node == null) {
            node = new Node(key, value);
            cache.put(key, node);
            addToHead(node);
            size++;
            if (size > capacity) {
                Node tail = removeTail();
                cache.remove(tail.key);
                size--;
            }
        }
        node.value = value;
        moveToHead(node);
    }

    private void addToHead(Node node) {
        node.prev = head;
        node.next = head.next;
        head.next.prev = node;
        head.next = node;
    }

    private void removeNode(Node node) {
        node.prev.next = node.next;
        node.next.prev = node.prev;
    }

    private void moveToHead(Node node) {
        removeNode(node);
        addToHead(node);
    }

    private Node removeTail() {
        Node res = tail.prev;
        removeNode(res);
        return res;
    }

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