mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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180 lines
29 KiB
JSON
180 lines
29 KiB
JSON
{
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"data": {
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"question": {
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"questionId": "146",
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"questionFrontendId": "146",
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"categoryTitle": "Algorithms",
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"boundTopicId": 1074,
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"title": "LRU Cache",
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"titleSlug": "lru-cache",
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"content": "<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>\n\n<p>Implement the <code>LRUCache</code> class:</p>\n\n<ul>\n\t<li><code>LRUCache(int capacity)</code> Initialize the LRU cache with <strong>positive</strong> size <code>capacity</code>.</li>\n\t<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>\n\t<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>\n</ul>\n\n<p>The functions <code data-stringify-type=\"code\">get</code> and <code data-stringify-type=\"code\">put</code> must each run in <code>O(1)</code> average time complexity.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n["LRUCache", "put", "put", "get", "put", "get", "put", "get", "get", "get"]\n[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]\n<strong>Output</strong>\n[null, null, null, 1, null, -1, null, -1, 3, 4]\n\n<strong>Explanation</strong>\nLRUCache lRUCache = new LRUCache(2);\nlRUCache.put(1, 1); // cache is {1=1}\nlRUCache.put(2, 2); // cache is {1=1, 2=2}\nlRUCache.get(1); // return 1\nlRUCache.put(3, 3); // LRU key was 2, evicts key 2, cache is {1=1, 3=3}\nlRUCache.get(2); // returns -1 (not found)\nlRUCache.put(4, 4); // LRU key was 1, evicts key 1, cache is {4=4, 3=3}\nlRUCache.get(1); // return -1 (not found)\nlRUCache.get(3); // return 3\nlRUCache.get(4); // return 4\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= capacity <= 3000</code></li>\n\t<li><code>0 <= key <= 10<sup>4</sup></code></li>\n\t<li><code>0 <= value <= 10<sup>5</sup></code></li>\n\t<li>At most 2<code> * 10<sup>5</sup></code> calls will be made to <code>get</code> and <code>put</code>.</li>\n</ul>\n",
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"translatedTitle": "LRU 缓存",
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"translatedContent": "<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>\n\n<div class=\"title__3Vvk\">实现 <code>LRUCache</code> 类：</div>\n\n<div class=\"original__bRMd\">\n<div>\n<ul>\n\t<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>\n\t<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>\n\t<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>\n</ul>\n\n<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>\n</div>\n</div>\n\n<p> </p>\n\n<p><strong>示例：</strong></p>\n\n<pre>\n<strong>输入</strong>\n[\"LRUCache\", \"put\", \"put\", \"get\", \"put\", \"get\", \"put\", \"get\", \"get\", \"get\"]\n[[2], [1, 1], [2, 2], [1], [3, 3], [2], [4, 4], [1], [3], [4]]\n<strong>输出</strong>\n[null, null, null, 1, null, -1, null, -1, 3, 4]\n\n<strong>解释</strong>\nLRUCache lRUCache = new LRUCache(2);\nlRUCache.put(1, 1); // 缓存是 {1=1}\nlRUCache.put(2, 2); // 缓存是 {1=1, 2=2}\nlRUCache.get(1); // 返回 1\nlRUCache.put(3, 3); // 该操作会使得关键字 2 作废，缓存是 {1=1, 3=3}\nlRUCache.get(2); // 返回 -1 (未找到)\nlRUCache.put(4, 4); // 该操作会使得关键字 1 作废，缓存是 {4=4, 3=3}\nlRUCache.get(1); // 返回 -1 (未找到)\nlRUCache.get(3); // 返回 3\nlRUCache.get(4); // 返回 4\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= capacity <= 3000</code></li>\n\t<li><code>0 <= key <= 10000</code></li>\n\t<li><code>0 <= value <= 10<sup>5</sup></code></li>\n\t<li>最多调用 <code>2 * 10<sup>5</sup></code> 次 <code>get</code> 和 <code>put</code></li>\n</ul>\n",
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"lang": "C++",
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"code": "class LRUCache {\npublic:\n LRUCache(int capacity) {\n\n }\n \n int get(int key) {\n\n }\n \n void put(int key, int value) {\n\n }\n};\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * LRUCache* obj = new LRUCache(capacity);\n * int param_1 = obj->get(key);\n * obj->put(key,value);\n */",
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"code": "class LRUCache {\n\n public LRUCache(int capacity) {\n\n }\n \n public int get(int key) {\n\n }\n \n public void put(int key, int value) {\n\n }\n}\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * LRUCache obj = new LRUCache(capacity);\n * int param_1 = obj.get(key);\n * obj.put(key,value);\n */",
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"code": "class LRUCache(object):\n\n def __init__(self, capacity):\n \"\"\"\n :type capacity: int\n \"\"\"\n\n\n def get(self, key):\n \"\"\"\n :type key: int\n :rtype: int\n \"\"\"\n\n\n def put(self, key, value):\n \"\"\"\n :type key: int\n :type value: int\n :rtype: None\n \"\"\"\n\n\n\n# Your LRUCache object will be instantiated and called as such:\n# obj = LRUCache(capacity)\n# param_1 = obj.get(key)\n# obj.put(key,value)",
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"code": "class LRUCache:\n\n def __init__(self, capacity: int):\n\n\n def get(self, key: int) -> int:\n\n\n def put(self, key: int, value: int) -> None:\n\n\n\n# Your LRUCache object will be instantiated and called as such:\n# obj = LRUCache(capacity)\n# param_1 = obj.get(key)\n# obj.put(key,value)",
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"code": "\n\n\ntypedef struct {\n\n} LRUCache;\n\n\nLRUCache* lRUCacheCreate(int capacity) {\n\n}\n\nint lRUCacheGet(LRUCache* obj, int key) {\n\n}\n\nvoid lRUCachePut(LRUCache* obj, int key, int value) {\n\n}\n\nvoid lRUCacheFree(LRUCache* obj) {\n\n}\n\n/**\n * Your LRUCache struct will be instantiated and called as such:\n * LRUCache* obj = lRUCacheCreate(capacity);\n * int param_1 = lRUCacheGet(obj, key);\n \n * lRUCachePut(obj, key, value);\n \n * lRUCacheFree(obj);\n*/",
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"code": "public class LRUCache {\n\n public LRUCache(int capacity) {\n\n }\n \n public int Get(int key) {\n\n }\n \n public void Put(int key, int value) {\n\n }\n}\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * LRUCache obj = new LRUCache(capacity);\n * int param_1 = obj.Get(key);\n * obj.Put(key,value);\n */",
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"lang": "JavaScript",
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"code": "/**\n * @param {number} capacity\n */\nvar LRUCache = function(capacity) {\n\n};\n\n/** \n * @param {number} key\n * @return {number}\n */\nLRUCache.prototype.get = function(key) {\n\n};\n\n/** \n * @param {number} key \n * @param {number} value\n * @return {void}\n */\nLRUCache.prototype.put = function(key, value) {\n\n};\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * var obj = new LRUCache(capacity)\n * var param_1 = obj.get(key)\n * obj.put(key,value)\n */",
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"code": "class LRUCache\n\n=begin\n :type capacity: Integer\n=end\n def initialize(capacity)\n\n end\n\n\n=begin\n :type key: Integer\n :rtype: Integer\n=end\n def get(key)\n\n end\n\n\n=begin\n :type key: Integer\n :type value: Integer\n :rtype: Void\n=end\n def put(key, value)\n\n end\n\n\nend\n\n# Your LRUCache object will be instantiated and called as such:\n# obj = LRUCache.new(capacity)\n# param_1 = obj.get(key)\n# obj.put(key, value)",
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"code": "\nclass LRUCache {\n\n init(_ capacity: Int) {\n\n }\n \n func get(_ key: Int) -> Int {\n\n }\n \n func put(_ key: Int, _ value: Int) {\n\n }\n}\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * let obj = LRUCache(capacity)\n * let ret_1: Int = obj.get(key)\n * obj.put(key, value)\n */",
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"code": "type LRUCache struct {\n\n}\n\n\nfunc Constructor(capacity int) LRUCache {\n\n}\n\n\nfunc (this *LRUCache) Get(key int) int {\n\n}\n\n\nfunc (this *LRUCache) Put(key int, value int) {\n\n}\n\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * obj := Constructor(capacity);\n * param_1 := obj.Get(key);\n * obj.Put(key,value);\n */",
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"code": "class LRUCache(_capacity: Int) {\n\n def get(key: Int): Int = {\n\n }\n\n def put(key: Int, value: Int) {\n\n }\n\n}\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * var obj = new LRUCache(capacity)\n * var param_1 = obj.get(key)\n * obj.put(key,value)\n */",
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"code": "class LRUCache(capacity: Int) {\n\n fun get(key: Int): Int {\n\n }\n\n fun put(key: Int, value: Int) {\n\n }\n\n}\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * var obj = LRUCache(capacity)\n * var param_1 = obj.get(key)\n * obj.put(key,value)\n */",
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"code": "struct LRUCache {\n\n}\n\n\n/**\n * `&self` means the method takes an immutable reference.\n * If you need a mutable reference, change it to `&mut self` instead.\n */\nimpl LRUCache {\n\n fn new(capacity: i32) -> Self {\n\n }\n \n fn get(&self, key: i32) -> i32 {\n\n }\n \n fn put(&self, key: i32, value: i32) {\n\n }\n}\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * let obj = LRUCache::new(capacity);\n * let ret_1: i32 = obj.get(key);\n * obj.put(key, value);\n */",
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"code": "class LRUCache {\n /**\n * @param Integer $capacity\n */\n function __construct($capacity) {\n\n }\n\n /**\n * @param Integer $key\n * @return Integer\n */\n function get($key) {\n\n }\n\n /**\n * @param Integer $key\n * @param Integer $value\n * @return NULL\n */\n function put($key, $value) {\n\n }\n}\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * $obj = LRUCache($capacity);\n * $ret_1 = $obj->get($key);\n * $obj->put($key, $value);\n */",
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"code": "class LRUCache {\n constructor(capacity: number) {\n\n }\n\n get(key: number): number {\n\n }\n\n put(key: number, value: number): void {\n\n }\n}\n\n/**\n * Your LRUCache object will be instantiated and called as such:\n * var obj = new LRUCache(capacity)\n * var param_1 = obj.get(key)\n * obj.put(key,value)\n */",
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"code": "(define lru-cache%\n (class object%\n (super-new)\n\n ; capacity : exact-integer?\n (init-field\n capacity)\n \n ; get : exact-integer? -> exact-integer?\n (define/public (get key)\n\n )\n ; put : exact-integer? exact-integer? -> void?\n (define/public (put key value)\n\n )))\n\n;; Your lru-cache% object will be instantiated and called as such:\n;; (define obj (new lru-cache% [capacity capacity]))\n;; (define param_1 (send obj get key))\n;; (send obj put key value)",
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"lang": "Erlang",
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"code": "-spec lru_cache_init_(Capacity :: integer()) -> any().\nlru_cache_init_(Capacity) ->\n .\n\n-spec lru_cache_get(Key :: integer()) -> integer().\nlru_cache_get(Key) ->\n .\n\n-spec lru_cache_put(Key :: integer(), Value :: integer()) -> any().\nlru_cache_put(Key, Value) ->\n .\n\n\n%% Your functions will be called as such:\n%% lru_cache_init_(Capacity),\n%% Param_1 = lru_cache_get(Key),\n%% lru_cache_put(Key, Value),\n\n%% lru_cache_init_ will be called before every test case, in which you can do some necessary initializations.",
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"code": "defmodule LRUCache do\n @spec init_(capacity :: integer) :: any\n def init_(capacity) do\n\n end\n\n @spec get(key :: integer) :: integer\n def get(key) do\n\n end\n\n @spec put(key :: integer, value :: integer) :: any\n def put(key, value) do\n\n end\nend\n\n# Your functions will be called as such:\n# LRUCache.init_(capacity)\n# param_1 = LRUCache.get(key)\n# LRUCache.put(key, value)\n\n# LRUCache.init_ will be called before every test case, in which you can do some necessary initializations.",
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