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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>

LRU Cache

LFU Cache

lfu-cache

LFU 缓存

Design In-Memory File System

design-in-memory-file-system

设计内存文件系统

Design Compressed String Iterator

design-compressed-string-iterator

迭代压缩字符串

Design a Load Distributor

design-a-load-distributor

Design Most Recently Used Queue

design-most-recently-used-queue

设计最近使用（MRU）队列

Design

Hash Table

Linked List

Doubly-Linked List

<div class="title__3Vvk">请你设计并实现一个满足&nbsp; <a href="https://baike.baidu.com/item/LRU" target="_blank">LRU (最近最少使用) 缓存</a> 约束的数据结构。</div>

<div class="title__3Vvk">实现 <code>LRUCache</code> 类：</div>

<div class="original__bRMd">
<div>
<ul>
	<li><code>LRUCache(int capacity)</code> 以 正整数 作为容量&nbsp;<code>capacity</code> 初始化 LRU 缓存</li>
	<li><code>int get(int key)</code> 如果关键字 <code>key</code> 存在于缓存中，则返回关键字的值，否则返回 <code>-1</code> 。</li>
	<li><code>void put(int key, int value)</code>&nbsp;如果关键字&nbsp;<code>key</code> 已经存在，则变更其数据值&nbsp;<code>value</code> ；如果不存在，则向缓存中插入该组&nbsp;<code>key-value</code> 。如果插入操作导致关键字数量超过&nbsp;<code>capacity</code> ，则应该 逐出 最久未使用的关键字。</li>
</ul>

函数 <code>get</code> 和 <code>put</code> 必须以 <code>O(1)</code> 的平均时间复杂度运行。
</div>
</div>

&nbsp;

示例：

<pre>
输入
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
输出
[null, null, null, 1, null, -1, null, -1, 3, 4]

解释
LRUCache lRUCache = new LRUCache(2);
lRUCache.put(1, 1); // 缓存是 {1=1}
lRUCache.put(2, 2); // 缓存是 {1=1, 2=2}
lRUCache.get(1); // 返回 1
lRUCache.put(3, 3); // 该操作会使得关键字 2 作废，缓存是 {1=1, 3=3}
lRUCache.get(2); // 返回 -1 (未找到)
lRUCache.put(4, 4); // 该操作会使得关键字 1 作废，缓存是 {4=4, 3=3}
lRUCache.get(1); // 返回 -1 (未找到)
lRUCache.get(3); // 返回 3
lRUCache.get(4); // 返回 4
</pre>

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= capacity &lt;= 3000</code></li>
	<li><code>0 &lt;= key &lt;= 10000</code></li>
	<li><code>0 &lt;= value &lt;= 105</code></li>
	<li>最多调用 <code>2 * 105</code> 次 <code>get</code> 和 <code>put</code></li>
</ul>