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小墨
2023-12-09 18:42:21 +08:00
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leetcode/problem/lru-cache.html
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@@ -5,13 +5,13 @@
<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>
<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&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>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>Example 1:</strong></p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input</strong>
@@ -40,5 +40,5 @@ lRUCache.get(4); // return 4
<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>
<li>At most <code>2 * 10<sup>5</sup></code> calls will be made to <code>get</code> and <code>put</code>.</li>
</ul>