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{
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"question": {
"questionId": "442",
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"categoryTitle": "Algorithms",
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"title": "Find All Duplicates in an Array",
"titleSlug": "find-all-duplicates-in-an-array",
"content": "<p>Given an integer array <code>nums</code> of length <code>n</code> where all the integers of <code>nums</code> are in the range <code>[1, n]</code> and each integer appears <strong>at most</strong> <strong>twice</strong>, return <em>an array of all the integers that appears <strong>twice</strong></em>.</p>\n\n<p>You must write an algorithm that runs in <code>O(n)</code> time and uses only <em>constant</em> auxiliary space, excluding the space needed to store the output</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<pre><strong>Input:</strong> nums = [4,3,2,7,8,2,3,1]\n<strong>Output:</strong> [2,3]\n</pre><p><strong class=\"example\">Example 2:</strong></p>\n<pre><strong>Input:</strong> nums = [1,1,2]\n<strong>Output:</strong> [1]\n</pre><p><strong class=\"example\">Example 3:</strong></p>\n<pre><strong>Input:</strong> nums = [1]\n<strong>Output:</strong> []\n</pre>\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= n</code></li>\n\t<li>Each element in <code>nums</code> appears <strong>once</strong> or <strong>twice</strong>.</li>\n</ul>\n",
"translatedTitle": "数组中重复的数据",
"translatedContent": "<p>给你一个长度为 <code>n</code> 的整数数组 <code>nums</code> ,其中 <code>nums</code> 的所有整数都在范围 <code>[1, n]</code> 内,且每个整数出现 <strong>最多</strong><strong>两次</strong> 。请你找出所有出现 <strong>两次</strong> 的整数,并以数组形式返回。</p>\n\n<p>你必须设计并实现一个时间复杂度为 <code>O(n)</code> 且仅使用常量额外空间(不包括存储输出所需的空间)的算法解决此问题。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = [4,3,2,7,8,2,3,1]\n<strong>输出:</strong>[2,3]\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = [1,1,2]\n<strong>输出:</strong>[1]\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = [1]\n<strong>输出:</strong>[]\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>1 &lt;= nums[i] &lt;= n</code></li>\n\t<li><code>nums</code> 中的每个元素出现 <strong>一次</strong> 或 <strong>两次</strong></li>\n</ul>\n",
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