LRU Cache

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.

Follow up: Could you do both operations in O(1) time complexity?

Example:

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

用单链表 模拟。只要动了，。就把他移到最后

map记录的是要记录的node的key和这个node之前的那个node。已便于移动，

class LRUCache {
    
    class Node{
        public int key, val;
        
        public Node next;
        public Node(int key, int value){
            this.key = key;
            this.val = value;
            this.next = null;
        }
    }
    
    private int capacity, size;
    private Node dummy,tail;
    private Map<Integer,Node> map;
    public LRUCache(int capacity) {
        this.capacity = capacity;
        size = 0;
        map = new HashMap<>();
        
        this.dummy = new Node(0,0);
        tail = dummy;
    }
    
    public void moveToTail(int key){
        Node prev = map.get(key);
        
        Node cur = prev.next;
        
        if(tail == cur){
            return;
        }
        
        //cur != tail. 说明prev的next的next不为空
        
        prev.next = prev.next.next;
        
        tail.next = cur;
        
        
        cur.next = null;
        
        if(prev.next != null)
            map.put(prev.next.key,prev);
        
        map.put(key,tail);
        tail = cur;
        
        
    }
    
    public int get(int key) {
        if(!map.containsKey(key)){
            return -1;
        }
        
        moveToTail(key);
        return tail.val;
    }
    
    public void put(int key, int value) {
        //get method move key to end 
        if(get(key) != -1){
            Node prev = map.get(key);
            prev.next.val = value;
            return;
        }
        
        if(size < capacity){
            Node n = new Node(key,value);
            
            size++;
            
            tail.next = n;
            
            map.put(key,tail);
            
            tail = n;
            return;
        }
        
        
        Node LRU = dummy.next;
        
        map.remove(LRU.key);
        
        LRU.key = key;
        LRU.val = value;
        
        map.put(key,dummy);
        moveToTail(key);
        
        
    }
    
    
}

/**
 * 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);
 */