Nav: << previous: 145.二叉树的后序遍历 | next: 147.对链表进行插入排序 >>
Description
tab: English
<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> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input</strong>
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[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> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= capacity <= 3000</code></li>
<li><code>0 <= key <= 10<sup>4</sup></code></li>
<li><code>0 <= value <= 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>
---
[submissions](https://leetcode.com/problems/lru-cache/submissions/) | [solutions](https://leetcode.com/problems/lru-cache/solutions/)
tab: 中文
<div class="title__3Vvk">请你设计并实现一个满足 <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 <strong>正整数</strong> 作为容量 <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> 如果关键字 <code>key</code> 已经存在，则变更其数据值 <code>value</code> ；如果不存在，则向缓存中插入该组 <code>key-value</code> 。如果插入操作导致关键字数量超过 <code>capacity</code> ，则应该 <strong>逐出</strong> 最久未使用的关键字。</li>
</ul>
<p>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。</p>
</div>
</div>
<p> </p>
<p><strong>示例：</strong></p>
<pre>
<strong>输入</strong>
["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]
[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]
<strong>输出</strong>
[null, null, null, 1, null, -1, null, -1, 3, 4]
<strong>解释</strong>
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>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>1 <= capacity <= 3000</code></li>
<li><code>0 <= key <= 10000</code></li>
<li><code>0 <= value <= 10<sup>5</sup></code></li>
<li>最多调用 <code>2 * 10<sup>5</sup></code> 次 <code>get</code> 和 <code>put</code></li>
</ul>
---
[提交记录](https://leetcode.cn/problems/lru-cache/submissions/) | [题解](https://leetcode.cn/problems/lru-cache/solution/)
Solutions & Notes
properties:
note.updated:
displayName: Last Updated
note.relative_links:
displayName: Related Links
note.desc:
displayName: Description
note.grade:
displayName: Rating
note.program_language:
displayName: Language
note.time_complexity:
displayName: TC
note.space_complexity:
displayName: SC
views:
- type: table
name: Solutions & Notes
filters:
and:
- file.hasLink(this.file)
- file.tags.containsAny("leetcode/solution", "leetcode/note")
order:
- file.name
- desc
- program_language
- time_complexity
- space_complexity
- grade
- relative_links
- updated
sort:
- property: grade
direction: ASC
- property: time_complexity
direction: ASC
- property: program_language
direction: ASC
columnSize:
file.name: 104
note.space_complexity: 65
note.grade: 126
Similar Problems
properties:
note.lcTopics:
displayName: Topics
note.lcAcRate:
displayName: AC Rate
note.favorites:
displayName: Favorites
note.grade:
displayName: Rating
note.translatedTitle:
displayName: Title (CN)
note.lcDifficulty:
displayName: Difficulty
views:
- type: table
name: Similar Problems
filters:
and:
- file.hasLink(this.file)
- similarQuestions.contains(this.file)
order:
- file.name
- translatedTitle
- lcTopics
- lcDifficulty
- lcAcRate
- grade
- favorites
sort:
- property: file.name
direction: ASC
- property: lcTopics
direction: DESC
columnSize:
note.translatedTitle: 240
note.lcTopics: 347
note.lcAcRate: 75
note.grade: 122