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leetcode-problemset/leetcode-cn/problem (Chinese)/前 K 个高频元素 [top-k-frequent-elements].html
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<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code> ,请你返回其中出现频率前 <code>k</code> 高的元素。你可以按 <strong>任意顺序</strong> 返回答案。</p>
<p>&nbsp;</p>
<p><strong class="example">示例 1</strong></p>
<div class="example-block">
<p><span class="example-io"><b>输入:</b>nums = [1,1,1,2,2,3], k = 2</span></p>
<p><strong>输出:</strong><span class="example-io">[1,2]</span></p>
</div>
<p><strong class="example">示例 2</strong></p>
<div class="example-block">
<p><span class="example-io"><b>输入:</b>nums = [1], k = 1</span></p>
<p><span class="example-io"><b>输出:</b>[1]</span></p>
</div>
<p><strong class="example">示例 3</strong></p>
<div class="example-block">
<p><span class="example-io"><b>输入:</b>nums = [1,2,1,2,1,2,3,1,3,2], k = 2</span></p>
<p><strong>输出:</strong><span class="example-io">[1,2]</span></p>
</div>
<p>&nbsp;</p>
<p><strong>提示:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
<li><code>k</code> 的取值范围是 <code>[1, 数组中不相同的元素的个数]</code></li>
<li>题目数据保证答案唯一,换句话说,数组中前 <code>k</code> 个高频元素的集合是唯一的</li>
</ul>
<p>&nbsp;</p>
<p><strong>进阶:</strong>你所设计算法的时间复杂度 <strong>必须</strong> 优于 <code>O(n log n)</code> ,其中 <code>n</code><em>&nbsp;</em>是数组大小。</p>