mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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210 lines
23 KiB
JSON
210 lines
23 KiB
JSON
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"title": "Top K Frequent Elements",
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"content": "<p>Given an integer array <code>nums</code> and an integer <code>k</code>, return <em>the</em> <code>k</code> <em>most frequent elements</em>. You may return the answer in <strong>any order</strong>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<pre><strong>Input:</strong> nums = [1,1,1,2,2,3], k = 2\n<strong>Output:</strong> [1,2]\n</pre><p><strong class=\"example\">Example 2:</strong></p>\n<pre><strong>Input:</strong> nums = [1], k = 1\n<strong>Output:</strong> [1]\n</pre>\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>\n\t<li><code>-10<sup>4</sup> <= nums[i] <= 10<sup>4</sup></code></li>\n\t<li><code>k</code> is in the range <code>[1, the number of unique elements in the array]</code>.</li>\n\t<li>It is <strong>guaranteed</strong> that the answer is <strong>unique</strong>.</li>\n</ul>\n\n<p> </p>\n<p><strong>Follow up:</strong> Your algorithm's time complexity must be better than <code>O(n log n)</code>, where n is the array's size.</p>\n",
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"translatedTitle": "前 K 个高频元素",
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"translatedContent": "<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>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入: </strong>nums = [1,1,1,2,2,3], k = 2\n<strong>输出: </strong>[1,2]\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入: </strong>nums = [1], k = 1\n<strong>输出: </strong>[1]</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>\n\t<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>\n\t<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>\n</ul>\n\n<p> </p>\n\n<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> </em>是数组大小。</p>\n",
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