Implement an 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.

The cache is initialized with a positive capacity.

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.put(4, 4);    // evicts key 1
cache.get(3);       // returns 3
cache.get(4);       // returns 4```

Solution – Java Program

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

class LRUCache {

Node tail = new Node(0,0);
int capacity;
HashMap<Integer,Node> map = new HashMap<Integer,Node>();
public LRUCache(int capacity) {
this.capacity = capacity;
}

public int get(int key) {
if(map.containsKey(key))
{
Node node = map.get(key);
remove(node);
insert(node);
return node.value;
}
else
return -1;
}

public void put(int key, int value) {
if(map.containsKey(key)){
remove(map.get(key));
}
if(map.size() == capacity){
remove(tail.prev);
}
insert(new Node(key,value));
}

public void remove(Node node)
{
map.remove(node.key);
node.prev.next = node.next;
node.next.prev = node.prev;
}

public void insert(Node node)
{
map.put(node.key,node);