Java 实现 LRUCache

Java 实现 LRUCache

leetcode 146. LRU缓存

• 双向链表+HashMap

• 使用双向链表的原因：删除链表特定节点（将某个节点移动到链表头部 或 删除 尾节点都需要用到）需要获得其前驱，要在O(1)时间复杂度内完成则可使用双向链表

• 使用双向链表存key 和 value ：删除链表尾节点时，也需要删除HashMap对应记录，需要用到key，cache.remove(tail.key)

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

        public Node() {
        }

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

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

    public MyLRUCache(int capacity) {
        this.size = 0;
        this.capacity = capacity;
        head = new Node();
        tail = new Node();
        head.next = tail;
        tail.prev = head;
    }

    public void put(int key, int value) {
        Node node = cache.get(key);
        if (node != null) {
            node.value = value;
            deleteNode(node);
            addToHead(node);
        } else {
            Node newNode = new Node(key, value);
            cache.put(key, newNode);
            addToHead(newNode);
            size++;
            if (size > capacity) {
                Node tail = deleteTail();
                cache.remove(tail.key);
                size--;
            }
        }
    }

    private Node deleteTail() {
        Node last = tail.prev;
        deleteNode(last);
        return last;
    }

    public int get(int key) {
        Node node = cache.get(key);
        if (node == null) {
            return -1;
        }
        deleteNode(node);
        addToHead(node);
        return node.value;
    }

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

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

}