leetcode-problemset/leetcode-cn/originData/majority-element.json

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            "categoryTitle": "Algorithms",
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            "title": "Majority Element",
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            "content": "<p>Given an array <code>nums</code> of size <code>n</code>, return <em>the majority element</em>.</p>\n\n<p>The majority element is the element that appears more than <code>&lfloor;n / 2&rfloor;</code> times. You may assume that the majority element always exists in the array.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<pre><strong>Input:</strong> nums = [3,2,3]\n<strong>Output:</strong> 3\n</pre><p><strong class=\"example\">Example 2:</strong></p>\n<pre><strong>Input:</strong> nums = [2,2,1,1,1,2,2]\n<strong>Output:</strong> 2\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;= 5 * 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n<strong>Follow-up:</strong> Could you solve the problem in linear time and in <code>O(1)</code> space?",
            "translatedTitle": "多数元素",
            "translatedContent": "<p>给定一个大小为 <code>n</code><em> </em>的数组&nbsp;<code>nums</code> ，返回其中的多数元素。多数元素是指在数组中出现次数 <strong>大于</strong>&nbsp;<code>⌊ n/2 ⌋</code>&nbsp;的元素。</p>\n\n<p>你可以假设数组是非空的，并且给定的数组总是存在多数元素。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例&nbsp;1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3,2,3]\n<strong>输出：</strong>3</pre>\n\n<p><strong>示例&nbsp;2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,2,1,1,1,2,2]\n<strong>输出：</strong>2\n</pre>\n\n<p>&nbsp;</p>\n<strong>提示：</strong>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>尝试设计时间复杂度为 O(n)、空间复杂度为 O(1) 的算法解决此问题。</p>\n",
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