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Algorithms

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

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

<ul>
	<li><code>LRUCache(int capacity)</code> Initialize the LRU cache with positive 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, evict the least recently used key.</li>
</ul>

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

&nbsp;
Example 1:

<pre>
Input
[&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]]
Output
[null, null, null, 1, null, -1, null, -1, 3, 4]

Explanation
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>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= capacity &lt;= 3000</code></li>
	<li><code>0 &lt;= key &lt;= 104</code></li>
	<li><code>0 &lt;= value &lt;= 105</code></li>
	<li>At most <code>2 * 105</code> calls will be made to <code>get</code> and <code>put</code>.</li>
</ul>