leetcode-problemset/leetcode-cn/originData/design-memory-allocator.json

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    "data": {
        "question": {
            "questionId": "2587",
            "questionFrontendId": "2502",
            "categoryTitle": "Algorithms",
            "boundTopicId": 2012709,
            "title": "Design Memory Allocator",
            "titleSlug": "design-memory-allocator",
            "content": "<p>You are given an integer <code>n</code> representing the size of a <strong>0-indexed</strong> memory array. All memory units are initially free.</p>\n\n<p>You have a memory allocator with the following functionalities:</p>\n\n<ol>\n\t<li><strong>Allocate </strong>a block of <code>size</code> consecutive free memory units and assign it the id <code>mID</code>.</li>\n\t<li><strong>Free</strong> all memory units with the given id <code>mID</code>.</li>\n</ol>\n\n<p><strong>Note</strong> that:</p>\n\n<ul>\n\t<li>Multiple blocks can be allocated to the same <code>mID</code>.</li>\n\t<li>You should free all the memory units with <code>mID</code>, even if they were allocated in different blocks.</li>\n</ul>\n\n<p>Implement the <code>Allocator</code> class:</p>\n\n<ul>\n\t<li><code>Allocator(int n)</code> Initializes an <code>Allocator</code> object with a memory array of size <code>n</code>.</li>\n\t<li><code>int allocate(int size, int mID)</code> Find the <strong>leftmost</strong> block of <code>size</code> <strong>consecutive</strong> free memory units and allocate it with the id <code>mID</code>. Return the block&#39;s first index. If such a block does not exist, return <code>-1</code>.</li>\n\t<li><code>int free(int mID)</code> Free all memory units with the id <code>mID</code>. Return the number of memory units you have freed.</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n[&quot;Allocator&quot;, &quot;allocate&quot;, &quot;allocate&quot;, &quot;allocate&quot;, &quot;free&quot;, &quot;allocate&quot;, &quot;allocate&quot;, &quot;allocate&quot;, &quot;free&quot;, &quot;allocate&quot;, &quot;free&quot;]\n[[10], [1, 1], [1, 2], [1, 3], [2], [3, 4], [1, 1], [1, 1], [1], [10, 2], [7]]\n<strong>Output</strong>\n[null, 0, 1, 2, 1, 3, 1, 6, 3, -1, 0]\n\n<strong>Explanation</strong>\nAllocator loc = new Allocator(10); // Initialize a memory array of size 10. All memory units are initially free.\nloc.allocate(1, 1); // The leftmost block&#39;s first index is 0. The memory array becomes [<strong>1</strong>,_,_,_,_,_,_,_,_,_]. We return 0.\nloc.allocate(1, 2); // The leftmost block&#39;s first index is 1. The memory array becomes [1,<strong>2</strong>,_,_,_,_,_,_,_,_]. We return 1.\nloc.allocate(1, 3); // The leftmost block&#39;s first index is 2. The memory array becomes [1,2,<strong>3</strong>,_,_,_,_,_,_,_]. We return 2.\nloc.free(2); // Free all memory units with mID 2. The memory array becomes [1,_, 3,_,_,_,_,_,_,_]. We return 1 since there is only 1 unit with mID 2.\nloc.allocate(3, 4); // The leftmost block&#39;s first index is 3. The memory array becomes [1,_,3,<strong>4</strong>,<strong>4</strong>,<strong>4</strong>,_,_,_,_]. We return 3.\nloc.allocate(1, 1); // The leftmost block&#39;s first index is 1. The memory array becomes [1,<strong>1</strong>,3,4,4,4,_,_,_,_]. We return 1.\nloc.allocate(1, 1); // The leftmost block&#39;s first index is 6. The memory array becomes [1,1,3,4,4,4,<strong>1</strong>,_,_,_]. We return 6.\nloc.free(1); // Free all memory units with mID 1. The memory array becomes [_,_,3,4,4,4,_,_,_,_]. We return 3 since there are 3 units with mID 1.\nloc.allocate(10, 2); // We can not find any free block with 10 consecutive free memory units, so we return -1.\nloc.free(7); // Free all memory units with mID 7. The memory array remains the same since there is no memory unit with mID 7. We return 0.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n, size, mID &lt;= 1000</code></li>\n\t<li>At most <code>1000</code> calls will be made to <code>allocate</code> and <code>free</code>.</li>\n</ul>\n",
            "translatedTitle": "设计内存分配器",
            "translatedContent": "<p>给你一个整数 <code>n</code> ，表示下标从 <strong>0</strong> 开始的内存数组的大小。所有内存单元开始都是空闲的。</p>\n\n<p>请你设计一个具备以下功能的内存分配器：</p>\n\n<ol>\n\t<li><strong>分配 </strong>一块大小为 <code>size</code> 的连续空闲内存单元并赋 id <code>mID</code> 。</li>\n\t<li><strong>释放</strong> 给定 id <code>mID</code> 对应的所有内存单元。</li>\n</ol>\n\n<p><strong>注意：</strong></p>\n\n<ul>\n\t<li>多个块可以被分配到同一个 <code>mID</code> 。</li>\n\t<li>你必须释放 <code>mID</code> 对应的所有内存单元，即便这些内存单元被分配在不同的块中。</li>\n</ul>\n\n<p>实现 <code>Allocator</code> 类：</p>\n\n<ul>\n\t<li><code>Allocator(int n)</code> 使用一个大小为 <code>n</code> 的内存数组初始化 <code>Allocator</code> 对象。</li>\n\t<li><code>int allocate(int size, int mID)</code> 找出大小为 <code>size</code> 个连续空闲内存单元且位于&nbsp; <strong>最左侧</strong> 的块，分配并赋 id <code>mID</code> 。返回块的第一个下标。如果不存在这样的块，返回 <code>-1</code> 。</li>\n\t<li><code>int free(int mID)</code> 释放 id <code>mID</code> 对应的所有内存单元。返回释放的内存单元数目。</li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例：</strong></p>\n\n<pre><strong>输入</strong>\n[\"Allocator\", \"allocate\", \"allocate\", \"allocate\", \"free\", \"allocate\", \"allocate\", \"allocate\", \"free\", \"allocate\", \"free\"]\n[[10], [1, 1], [1, 2], [1, 3], [2], [3, 4], [1, 1], [1, 1], [1], [10, 2], [7]]\n<strong>输出</strong>\n[null, 0, 1, 2, 1, 3, 1, 6, 3, -1, 0]\n\n<strong>解释</strong>\nAllocator loc = new Allocator(10); // 初始化一个大小为 10 的内存数组，所有内存单元都是空闲的。\nloc.allocate(1, 1); // 最左侧的块的第一个下标是 0 。内存数组变为 [<strong>1</strong>, , , , , , , , , ]。返回 0 。\nloc.allocate(1, 2); // 最左侧的块的第一个下标是 1 。内存数组变为 [1,<strong>2</strong>, , , , , , , , ]。返回 1 。\nloc.allocate(1, 3); // 最左侧的块的第一个下标是 2 。内存数组变为 [1,2,<strong>3</strong>, , , , , , , ]。返回 2 。\nloc.free(2); // 释放 mID 为 2 的所有内存单元。内存数组变为 [1, ,<strong>3</strong>, , , , , , , ] 。返回 1 ，因为只有 1 个 mID 为 2 的内存单元。\nloc.allocate(3, 4); // 最左侧的块的第一个下标是 3 。内存数组变为 [1, ,3,<strong>4</strong>,<strong>4</strong>,<strong>4</strong>, , , , ]。返回 3 。\nloc.allocate(1, 1); // 最左侧的块的第一个下标是 1 。内存数组变为 [1,<strong>1</strong>,3,4,4,4, , , , ]。返回 1 。\nloc.allocate(1, 1); // 最左侧的块的第一个下标是 6 。内存数组变为 [1,1,3,4,4,4,<strong>1</strong>, , , ]。返回 6 。\nloc.free(1); // 释放 mID 为 1 的所有内存单元。内存数组变为 [ , ,3,4,4,4,<strong> </strong>, , , ] 。返回 3 ，因为有 3 个 mID 为 1 的内存单元。\nloc.allocate(10, 2); // 无法找出长度为 10 个连续空闲内存单元的空闲块，所有返回 -1 。\nloc.free(7); // 释放 mID 为 7 的所有内存单元。内存数组保持原状，因为不存在 mID 为 7 的内存单元。返回 0 。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= n, size, mID &lt;= 1000</code></li>\n\t<li>最多调用 <code>allocate</code> 和 <code>free</code> 方法 <code>1000</code> 次</li>\n</ul>\n",
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                    "lang": "C++",
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                    "code": "class Allocator {\npublic:\n    Allocator(int n) {\n\n    }\n    \n    int allocate(int size, int mID) {\n\n    }\n    \n    int free(int mID) {\n\n    }\n};\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * Allocator* obj = new Allocator(n);\n * int param_1 = obj->allocate(size,mID);\n * int param_2 = obj->free(mID);\n */",
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                    "code": "class Allocator {\n\n    public Allocator(int n) {\n\n    }\n    \n    public int allocate(int size, int mID) {\n\n    }\n    \n    public int free(int mID) {\n\n    }\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * Allocator obj = new Allocator(n);\n * int param_1 = obj.allocate(size,mID);\n * int param_2 = obj.free(mID);\n */",
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                    "code": "class Allocator(object):\n\n    def __init__(self, n):\n        \"\"\"\n        :type n: int\n        \"\"\"\n\n\n    def allocate(self, size, mID):\n        \"\"\"\n        :type size: int\n        :type mID: int\n        :rtype: int\n        \"\"\"\n\n\n    def free(self, mID):\n        \"\"\"\n        :type mID: int\n        :rtype: int\n        \"\"\"\n\n\n\n# Your Allocator object will be instantiated and called as such:\n# obj = Allocator(n)\n# param_1 = obj.allocate(size,mID)\n# param_2 = obj.free(mID)",
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                    "code": "class Allocator:\n\n    def __init__(self, n: int):\n\n\n    def allocate(self, size: int, mID: int) -> int:\n\n\n    def free(self, mID: int) -> int:\n\n\n\n# Your Allocator object will be instantiated and called as such:\n# obj = Allocator(n)\n# param_1 = obj.allocate(size,mID)\n# param_2 = obj.free(mID)",
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                    "code": "\n\n\ntypedef struct {\n    \n} Allocator;\n\n\nAllocator* allocatorCreate(int n) {\n    \n}\n\nint allocatorAllocate(Allocator* obj, int size, int mID) {\n    \n}\n\nint allocatorFree(Allocator* obj, int mID) {\n    \n}\n\nvoid allocatorFree(Allocator* obj) {\n    \n}\n\n/**\n * Your Allocator struct will be instantiated and called as such:\n * Allocator* obj = allocatorCreate(n);\n * int param_1 = allocatorAllocate(obj, size, mID);\n \n * int param_2 = allocatorFree(obj, mID);\n \n * allocatorFree(obj);\n*/",
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                    "lang": "C#",
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                    "code": "public class Allocator {\n\n    public Allocator(int n) {\n\n    }\n    \n    public int Allocate(int size, int mID) {\n\n    }\n    \n    public int Free(int mID) {\n\n    }\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * Allocator obj = new Allocator(n);\n * int param_1 = obj.Allocate(size,mID);\n * int param_2 = obj.Free(mID);\n */",
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                    "code": "/**\n * @param {number} n\n */\nvar Allocator = function(n) {\n\n};\n\n/** \n * @param {number} size \n * @param {number} mID\n * @return {number}\n */\nAllocator.prototype.allocate = function(size, mID) {\n\n};\n\n/** \n * @param {number} mID\n * @return {number}\n */\nAllocator.prototype.free = function(mID) {\n\n};\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * var obj = new Allocator(n)\n * var param_1 = obj.allocate(size,mID)\n * var param_2 = obj.free(mID)\n */",
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                    "lang": "TypeScript",
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                    "code": "class Allocator {\n    constructor(n: number) {\n        \n    }\n\n    allocate(size: number, mID: number): number {\n        \n    }\n\n    free(mID: number): number {\n        \n    }\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * var obj = new Allocator(n)\n * var param_1 = obj.allocate(size,mID)\n * var param_2 = obj.free(mID)\n */",
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                    "code": "class Allocator {\n    /**\n     * @param Integer $n\n     */\n    function __construct($n) {\n\n    }\n\n    /**\n     * @param Integer $size\n     * @param Integer $mID\n     * @return Integer\n     */\n    function allocate($size, $mID) {\n\n    }\n\n    /**\n     * @param Integer $mID\n     * @return Integer\n     */\n    function free($mID) {\n\n    }\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * $obj = Allocator($n);\n * $ret_1 = $obj->allocate($size, $mID);\n * $ret_2 = $obj->free($mID);\n */",
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                    "code": "\nclass Allocator {\n\n    init(_ n: Int) {\n\n    }\n    \n    func allocate(_ size: Int, _ mID: Int) -> Int {\n\n    }\n    \n    func free(_ mID: Int) -> Int {\n\n    }\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * let obj = Allocator(n)\n * let ret_1: Int = obj.allocate(size, mID)\n * let ret_2: Int = obj.free(mID)\n */",
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                    "lang": "Kotlin",
                    "langSlug": "kotlin",
                    "code": "class Allocator(n: Int) {\n\n    fun allocate(size: Int, mID: Int): Int {\n\n    }\n\n    fun free(mID: Int): Int {\n\n    }\n\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * var obj = Allocator(n)\n * var param_1 = obj.allocate(size,mID)\n * var param_2 = obj.free(mID)\n */",
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                    "code": "class Allocator {\n\n  Allocator(int n) {\n    \n  }\n  \n  int allocate(int size, int mID) {\n    \n  }\n  \n  int free(int mID) {\n    \n  }\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * Allocator obj = Allocator(n);\n * int param1 = obj.allocate(size,mID);\n * int param2 = obj.free(mID);\n */",
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                    "code": "type Allocator struct {\n\n}\n\n\nfunc Constructor(n int) Allocator {\n\n}\n\n\nfunc (this *Allocator) Allocate(size int, mID int) int {\n\n}\n\n\nfunc (this *Allocator) Free(mID int) int {\n\n}\n\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * obj := Constructor(n);\n * param_1 := obj.Allocate(size,mID);\n * param_2 := obj.Free(mID);\n */",
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                    "lang": "Ruby",
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                    "code": "class Allocator\n\n=begin\n    :type n: Integer\n=end\n    def initialize(n)\n\n    end\n\n\n=begin\n    :type size: Integer\n    :type m_id: Integer\n    :rtype: Integer\n=end\n    def allocate(size, m_id)\n\n    end\n\n\n=begin\n    :type m_id: Integer\n    :rtype: Integer\n=end\n    def free(m_id)\n\n    end\n\n\nend\n\n# Your Allocator object will be instantiated and called as such:\n# obj = Allocator.new(n)\n# param_1 = obj.allocate(size, m_id)\n# param_2 = obj.free(m_id)",
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                    "lang": "Scala",
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                    "code": "class Allocator(_n: Int) {\n\n    def allocate(size: Int, mID: Int): Int = {\n\n    }\n\n    def free(mID: Int): Int = {\n\n    }\n\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * var obj = new Allocator(n)\n * var param_1 = obj.allocate(size,mID)\n * var param_2 = obj.free(mID)\n */",
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                    "code": "struct Allocator {\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 Allocator {\n\n    fn new(n: i32) -> Self {\n\n    }\n    \n    fn allocate(&self, size: i32, m_id: i32) -> i32 {\n\n    }\n    \n    fn free(&self, m_id: i32) -> i32 {\n\n    }\n}\n\n/**\n * Your Allocator object will be instantiated and called as such:\n * let obj = Allocator::new(n);\n * let ret_1: i32 = obj.allocate(size, mID);\n * let ret_2: i32 = obj.free(mID);\n */",
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                    "code": "(define allocator%\n  (class object%\n    (super-new)\n    \n    ; n : exact-integer?\n    (init-field\n      n)\n    \n    ; allocate : exact-integer? exact-integer? -> exact-integer?\n    (define/public (allocate size m-id)\n      )\n    ; free : exact-integer? -> exact-integer?\n    (define/public (free m-id)\n      )))\n\n;; Your allocator% object will be instantiated and called as such:\n;; (define obj (new allocator% [n n]))\n;; (define param_1 (send obj allocate size m-id))\n;; (define param_2 (send obj free m-id))",
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                    "lang": "Erlang",
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                    "code": "-spec allocator_init_(N :: integer()) -> any().\nallocator_init_(N) ->\n  .\n\n-spec allocator_allocate(Size :: integer(), MID :: integer()) -> integer().\nallocator_allocate(Size, MID) ->\n  .\n\n-spec allocator_free(MID :: integer()) -> integer().\nallocator_free(MID) ->\n  .\n\n\n%% Your functions will be called as such:\n%% allocator_init_(N),\n%% Param_1 = allocator_allocate(Size, MID),\n%% Param_2 = allocator_free(MID),\n\n%% allocator_init_ will be called before every test case, in which you can do some necessary initializations.",
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                    "code": "defmodule Allocator do\n  @spec init_(n :: integer) :: any\n  def init_(n) do\n    \n  end\n\n  @spec allocate(size :: integer, m_id :: integer) :: integer\n  def allocate(size, m_id) do\n    \n  end\n\n  @spec free(m_id :: integer) :: integer\n  def free(m_id) do\n    \n  end\nend\n\n# Your functions will be called as such:\n# Allocator.init_(n)\n# param_1 = Allocator.allocate(size, m_id)\n# param_2 = Allocator.free(m_id)\n\n# Allocator.init_ will be called before every test case, in which you can do some necessary initializations.",
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                "Can you simulate the process?",
                "Use brute force to find the leftmost free block and free each occupied memory unit"
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            "sampleTestCase": "[\"Allocator\",\"allocate\",\"allocate\",\"allocate\",\"free\",\"allocate\",\"allocate\",\"allocate\",\"free\",\"allocate\",\"free\"]\n[[10],[1,1],[1,2],[1,3],[2],[3,4],[1,1],[1,1],[1],[10,2],[7]]",
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