update

leetcode-cn/originData/find-minimum-in-rotated-sorted-array.json

@@ -12,7 +12,7 @@
             "translatedContent": "已知一个长度为 <code>n</code> 的数组，预先按照升序排列，经由 <code>1</code> 到 <code>n</code> 次 <strong>旋转</strong> 后，得到输入数组。例如，原数组 <code>nums = [0,1,2,4,5,6,7]</code> 在变化后可能得到：\n<ul>\n\t<li>若旋转 <code>4</code> 次，则可以得到 <code>[4,5,6,7,0,1,2]</code></li>\n\t<li>若旋转 <code>7</code> 次，则可以得到 <code>[0,1,2,4,5,6,7]</code></li>\n</ul>\n\n<p>注意，数组 <code>[a[0], a[1], a[2], ..., a[n-1]]</code> <strong>旋转一次</strong> 的结果为数组 <code>[a[n-1], a[0], a[1], a[2], ..., a[n-2]]</code> 。</p>\n\n<p>给你一个元素值 <strong>互不相同</strong> 的数组 <code>nums</code> ，它原来是一个升序排列的数组，并按上述情形进行了多次旋转。请你找出并返回数组中的 <strong>最小元素</strong> 。</p>\n\n<p>你必须设计一个时间复杂度为&nbsp;<code>O(log n)</code> 的算法解决此问题。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [3,4,5,1,2]\n<strong>输出：</strong>1\n<strong>解释：</strong>原数组为 [1,2,3,4,5] ，旋转 3 次得到输入数组。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [4,5,6,7,0,1,2]\n<strong>输出：</strong>0\n<strong>解释：</strong>原数组为 [0,1,2,4,5,6,7] ，旋转 4 次得到输入数组。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [11,13,15,17]\n<strong>输出：</strong>11\n<strong>解释：</strong>原数组为 [11,13,15,17] ，旋转 4 次得到输入数组。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>n == nums.length</code></li>\n\t<li><code>1 &lt;= n &lt;= 5000</code></li>\n\t<li><code>-5000 &lt;= nums[i] &lt;= 5000</code></li>\n\t<li><code>nums</code> 中的所有整数 <strong>互不相同</strong></li>\n\t<li><code>nums</code> 原来是一个升序排序的数组，并进行了 <code>1</code> 至 <code>n</code> 次旋转</li>\n</ul>\n",
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             "hints": [
                 "Array was originally in ascending order. Now that the array is rotated, there would be a point in the array where there is a small deflection from the increasing sequence. eg. The array would be something like [4, 5, 6, 7, 0, 1, 2].",
                 "You can divide the search space into two and see which direction to go.\r\nCan you think of an algorithm which has O(logN) search complexity?",