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"categoryTitle": "Algorithms",
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"title": "Find Peak Element",
"titleSlug": "find-peak-element",
"content": "<p>A peak element is an element that is strictly greater than its neighbors.</p>\n\n<p>Given an integer array <code>nums</code>, find a peak element, and return its index. If&nbsp;the array contains multiple peaks, return the index to <strong>any of the peaks</strong>.</p>\n\n<p>You may imagine that <code>nums[-1] = nums[n] = -&infin;</code>.</p>\n\n<p>You must write an algorithm that runs in&nbsp;<code>O(log n)</code> time.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3,1]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> 3 is a peak element and your function should return the index number 2.</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,1,3,5,6,4]\n<strong>Output:</strong> 5\n<strong>Explanation:</strong> Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>\n\t<li><code>nums[i] != nums[i + 1]</code> for all valid <code>i</code>.</li>\n</ul>\n",
"translatedTitle": "寻找峰值",
"translatedContent": "<p>峰值元素是指其值严格大于左右相邻值的元素。</p>\n\n<p>给你一个整数数组&nbsp;<code>nums</code>,找到峰值元素并返回其索引。数组可能包含多个峰值,在这种情况下,返回 <strong>任何一个峰值</strong> 所在位置即可。</p>\n\n<p>你可以假设&nbsp;<code>nums[-1] = nums[n] = -∞</code> 。</p>\n\n<p>你必须实现时间复杂度为 <code>O(log n)</code><em> </em>的算法来解决此问题。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = <code>[1,2,3,1]</code>\n<strong>输出:</strong>2\n<strong>解释:</strong>3 是峰值元素,你的函数应该返回其索引 2。</pre>\n\n<p><strong>示例&nbsp;2</strong></p>\n\n<pre>\n<strong>输入:</strong>nums = <code>[</code>1,2,1,3,5,6,4]\n<strong>输出:</strong>1 或 5 \n<strong>解释:</strong>你的函数可以返回索引 1其峰值元素为 2\n&nbsp; 或者返回索引 5 其峰值元素为 6。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 1000</code></li>\n\t<li><code>-2<sup>31</sup> &lt;= nums[i] &lt;= 2<sup>31</sup> - 1</code></li>\n\t<li>对于所有有效的 <code>i</code> 都有 <code>nums[i] != nums[i + 1]</code></li>\n</ul>\n",
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