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check-if-array-is-sorted-and-rotated.json
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{
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"title": "Check if Array Is Sorted and Rotated",
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"content": "<p>Given an array <code>nums</code>, return <code>true</code><em> if the array was originally sorted in non-decreasing order, then rotated <strong>some</strong> number of positions (including zero)</em>. Otherwise, return <code>false</code>.</p>\n\n<p>There may be <strong>duplicates</strong> in the original array.</p>\n\n<p><strong>Note:</strong> An array <code>A</code> rotated by <code>x</code> positions results in an array <code>B</code> of the same length such that <code>A[i] == B[(i+x) % A.length]</code>, where <code>%</code> is the modulo operation.</p>\n\n<p> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [3,4,5,1,2]\n<strong>Output:</strong> true\n<strong>Explanation:</strong> [1,2,3,4,5] is the original sorted array.\nYou can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2].\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [2,1,3,4]\n<strong>Output:</strong> false\n<strong>Explanation:</strong> There is no sorted array once rotated that can make nums.\n</pre>\n\n<p><strong>Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> nums = [1,2,3]\n<strong>Output:</strong> true\n<strong>Explanation:</strong> [1,2,3] is the original sorted array.\nYou can rotate the array by x = 0 positions (i.e. no rotation) to make nums.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 100</code></li>\n\t<li><code>1 <= nums[i] <= 100</code></li>\n</ul>\n",
"translatedTitle": "检查数组是否经排序和轮转得到",
"translatedContent": "<p>给你一个数组 <code>nums</code> 。<code>nums</code> 的源数组中，所有元素与 <code>nums</code> 相同，但按非递减顺序排列。</p>\n\n<p>如果 <code>nums</code> 能够由源数组轮转若干位置（包括 0 个位置）得到，则返回 <code>true</code><em> </em>；否则，返回 <code>false</code> 。</p>\n\n<p>源数组中可能存在 <strong>重复项</strong> 。</p>\n\n<p><strong>注意：</strong>我们称数组 <code>A</code> 在轮转 <code>x</code> 个位置后得到长度相同的数组 <code>B</code> ，当它们满足 <code>A[i] == B[(i+x) % A.length]</code> ，其中 <code>%</code> 为取余运算。</p>\n\n<p> </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>true\n<strong>解释：</strong>[1,2,3,4,5] 为有序的源数组。\n可以轮转 x = 3 个位置，使新数组从值为 3 的元素开始：[3,4,5,1,2] 。\n</pre>\n\n<p><strong>示例 2：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [2,1,3,4]\n<strong>输出：</strong>false\n<strong>解释：</strong>源数组无法经轮转得到 nums 。\n</pre>\n\n<p><strong>示例 3：</strong></p>\n\n<pre>\n<strong>输入：</strong>nums = [1,2,3]\n<strong>输出：</strong>true\n<strong>解释：</strong>[1,2,3] 为有序的源数组。\n可以轮转 x = 0 个位置（即不轮转）得到 nums 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= nums.length <= 100</code></li>\n\t<li><code>1 <= nums[i] <= 100</code></li>\n</ul>\n",
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