LRU Cache
<p>Design a data structure that follows the constraints of a <strong><a href="https://en.wikipedia.org/wiki/Cache_replacement_policies#LRU" target="_blank">Least Recently Used (LRU) cache</a></strong>.</p>

<p>Implement the <code>LRUCache</code> class:</p>

<ul>
	<li><code>LRUCache(int capacity)</code> Initialize the LRU cache with <strong>positive</strong> size <code>capacity</code>.</li>
	<li><code>int get(int key)</code> Return the value of the <code>key</code> if the key exists, otherwise return <code>-1</code>.</li>
	<li><code>void put(int key, int value)</code> Update the value of the <code>key</code> if the <code>key</code> exists. Otherwise, add the <code>key-value</code> pair to the cache. If the number of keys exceeds the <code>capacity</code> from this operation, <strong>evict</strong> the least recently used key.</li>
</ul>

<p>The functions <code>get</code> and <code>put</code> must each run in <code>O(1)</code> average time complexity.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input</strong>
[&quot;LRUCache&quot;, &quot;put&quot;, &quot;put&quot;, &quot;get&quot;, &quot;put&quot;, &quot;get&quot;, &quot;put&quot;, &quot;get&quot;, &quot;get&quot;, &quot;get&quot;]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
<strong>Output</strong>
[null, null, null, 1, null, -1, null, -1, 3, 4]

<strong>Explanation</strong>
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // cache is {1=1}
lRUCache.put(2, 2); // cache is {1=1, 2=2}
lRUCache.get(1);    // return 1
lRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}
lRUCache.get(2);    // returns -1 (not found)
lRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}
lRUCache.get(1);    // return -1 (not found)
lRUCache.get(3);    // return 3
lRUCache.get(4);    // return 4
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= capacity &lt;= 3000</code></li>
	<li><code>0 &lt;= key &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= value &lt;= 10<sup>5</sup></code></li>
	<li>At most <code>2 * 10<sup>5</sup></code> calls will be made to <code>get</code> and <code>put</code>.</li>
</ul>
Design; Hash Table; Linked List; Doubly-Linked List
type Node struct {
	key, val   int
	prev, next *Node
}

type LRUCache struct {
	size, capacity int
	head, tail     *Node
	cache          map[int]*Node
}

func Constructor(capacity int) LRUCache {
	head := &Node{}
	tail := &Node{}
	head.next = tail
	tail.prev = head
	return LRUCache{
		capacity: capacity,
		head:     head,
		tail:     tail,
		cache:    make(map[int]*Node),
	}
}

func (this *LRUCache) Get(key int) int {
	if node, exists := this.cache[key]; exists {
		this.removeNode(node)
		this.addToHead(node)
		return node.val
	}
	return -1
}

func (this *LRUCache) Put(key int, value int) {
	if node, exists := this.cache[key]; exists {
		this.removeNode(node)
		node.val = value
		this.addToHead(node)
	} else {
		node := &Node{key: key, val: value}
		this.cache[key] = node
		this.addToHead(node)
		if this.size++; this.size > this.capacity {
			node = this.tail.prev
			delete(this.cache, node.key)
			this.removeNode(node)
			this.size--
		}
	}
}

func (this *LRUCache) removeNode(node *Node) {
	node.prev.next = node.next
	node.next.prev = node.prev
}

func (this *LRUCache) addToHead(node *Node) {
	node.next = this.head.next
	node.prev = this.head
	this.head.next = node
	node.next.prev = node
}

/**
 * Your LRUCache object will be instantiated and called as such:
 * obj := Constructor(capacity);
 * param_1 := obj.Get(key);
 * obj.Put(key,value);
 */

/**
 * Your LRUCache object will be instantiated and called as such:
 * obj := Constructor(capacity);
 * param_1 := obj.Get(key);
 * obj.Put(key,value);
 */
