LeetCode 146. LRU 缓存

LeetCode 146. LRU 缓存

问题描述：

解题思路：

采用哈希表+双向链表较为方便时间。双向链表便于维护数据最近使用再前，最久未使用在后。双向链表便于移动删除。

题解：

class LRUCache {

    class DlinkedNode{
        int key;
        int value;
        DlinkedNode prev;
        DlinkedNode next;
        public DlinkedNode(){};
        public DlinkedNode(int _key,int _value){key = _key;value = _value;};
    }
    private Mapcache = new HashMap();
    private int size;
    private int capacity;
    private DlinkedNode head;
    private DlinkedNode tail;

    public LRUCache(int capacity) {
        this.capacity = capacity;
        this.size = 0;
        head = new DlinkedNode();
        tail = new DlinkedNode();
        head.next = tail;
        tail.prev = head;
    }
    
    public int get(int key) {
        DlinkedNode node = cache.get(key);
        if(node == null){
            return -1;
        }
        moveToHead(node);
        return node.value;
    }
    
    public void put(int key, int value) {
        DlinkedNode node = cache.get(key);
        if(node == null){
            DlinkedNode node1 = new DlinkedNode(key,value);
            cache.put(key,node1);
            addToHead(node1);
            size++;
            if(size>capacity){
                DlinkedNode node2 = removeToTail();
                cache.remove(node2.key);
                size--;
            }
        }else{
            node.value = value;
            moveToHead(node);
        }
    }
    private void addToHead(DlinkedNode node){
        node.prev = head;
        node.next = head.next;
        head.next.prev = node;
        head.next = node;
    }

    private void removeNode(DlinkedNode node){
        node.prev.next = node.next;
        node.next.prev = node.prev;
    }
    private void moveToHead(DlinkedNode node){
        removeNode(node);
        addToHead(node);
    }
    private DlinkedNode removeToTail(){
        DlinkedNode node = tail.prev;
        removeNode(node);
        return node;
    }
}