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<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>
<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>
<p>&nbsp;</p>
<p><strong>示例&nbsp;1</strong></p>
<pre>
<strong>输入:</strong>nums = [3,2,3]
<strong>输出:</strong>3</pre>
<p><strong>示例&nbsp;2</strong></p>
<pre>
<strong>输入:</strong>nums = [2,2,1,1,1,2,2]
<strong>输出:</strong>2
</pre>
<p>&nbsp;</p>
<strong>提示:</strong>
<ul>
<li><code>n == nums.length</code></li>
<li><code>1 &lt;= n &lt;= 5 * 10<sup>4</sup></code></li>
<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>
<p>&nbsp;</p>
<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>