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leetcode-problemset/leetcode/problem/lru-cache.html
2022-03-29 12:55:24 +08:00

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<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>&nbsp;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&nbsp;<code data-stringify-type="code">get</code>&nbsp;and&nbsp;<code data-stringify-type="code">put</code>&nbsp;must each run in <code>O(1)</code> average time complexity.</p>
<p>&nbsp;</p>
<p><strong>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 2<code>&nbsp;* 10<sup>5</sup></code>&nbsp;calls will be made to <code>get</code> and <code>put</code>.</li>
</ul>