leetcode-problemset/leetcode-cn/originData/counter.json

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        "question": {
            "questionId": "2732",
            "questionFrontendId": "2620",
            "categoryTitle": "JavaScript",
            "boundTopicId": 2222282,
            "title": "Counter",
            "titleSlug": "counter",
            "content": "<p>Given an integer&nbsp;<code>n</code>,&nbsp;return a <code>counter</code> function. This <code>counter</code> function initially returns&nbsp;<code>n</code>&nbsp;and then returns 1 more than the previous value every subsequent time it is called (<code>n</code>, <code>n + 1</code>, <code>n + 2</code>, etc).</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nn = 10 \n[&quot;call&quot;,&quot;call&quot;,&quot;call&quot;]\n<strong>Output:</strong> [10,11,12]\n<strong>Explanation: \n</strong>counter() = 10 // The first time counter() is called, it returns n.\ncounter() = 11 // Returns 1 more than the previous time.\ncounter() = 12 // Returns 1 more than the previous time.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> \nn = -2\n[&quot;call&quot;,&quot;call&quot;,&quot;call&quot;,&quot;call&quot;,&quot;call&quot;]\n<strong>Output:</strong> [-2,-1,0,1,2]\n<strong>Explanation:</strong> counter() initially returns -2. Then increases after each sebsequent call.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>-1000<sup>&nbsp;</sup>&lt;= n &lt;= 1000</code></li>\n\t<li><code>At most 1000 calls to counter() will be made</code></li>\n</ul>\n",
            "translatedTitle": "计数器",
            "translatedContent": "<p>请你编写并返回一个&nbsp;<strong>计数器&nbsp;</strong>函数，它接收一个整型参数 n 。这个&nbsp;<strong>计数器&nbsp;</strong>函数最初返回 n，每次调用它时返回前一个值加 1 的值 ( <code>n</code> ,&nbsp; <code>n + 1</code> ,&nbsp; <code>n + 2</code> ，等等)。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>\nn = 10 \n[\"call\",\"call\",\"call\"]\n<b>输出：</b>[10,11,12]\n<strong>解释：\n</strong>counter() = 10 // 第一次调用 counter()，返回 n。\ncounter() = 11 // 返回上次调用的值加 1。\ncounter() = 12 // 返回上次调用的值加 1。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>\nn = -2\n[\"call\",\"call\",\"call\",\"call\",\"call\"]\n<b>输出：</b>[-2,-1,0,1,2]\n<b>解释：</b>counter() 最初返回 -2。然后在每个后续调用后增加 1。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>-1000<sup>&nbsp;</sup>&lt;= n &lt;= 1000</code></li>\n\t<li><code>最多对 counter() 进行 1000 次调用</code></li>\n</ul>\n",
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            "difficulty": "Easy",
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                {
                    "lang": "JavaScript",
                    "langSlug": "javascript",
                    "code": "/**\n * @param {number} n\n * @return {Function} counter\n */\nvar createCounter = function(n) {\n    return function() {\n        \n    };\n};\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */",
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                {
                    "lang": "TypeScript",
                    "langSlug": "typescript",
                    "code": "function createCounter(n: number): () => number {\n    return function() {\n\n    }\n}\n\n\n/** \n * const counter = createCounter(10)\n * counter() // 10\n * counter() // 11\n * counter() // 12\n */",
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            "hints": [
                "In Javascript, a function can return a clojure. A clojure is defined as a function that can access variables declared above it (it's lexical environment).",
                "A count variable can be initialized in the outer function and mutated in the inner function."
            ],
            "solution": null,
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            "sampleTestCase": "10\n[\"call\",\"call\",\"call\"]",
            "metaData": "{\n  \"name\": \"createCounter\",\n  \"params\": [\n    {\n      \"name\": \"n\",\n      \"type\": \"integer\"\n    },\n    {\n      \"type\": \"string[]\",\n      \"name\": \"calls\"\n    }\n  ],\n  \"return\": {\n    \"type\": \"integer\"\n  },\n  \"languages\": [\n    \"javascript\",\n    \"typescript\"\n  ],\n  \"manual\": true\n}",
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