Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:
[4,5,6,7,0,1,4] if it was rotated 4 times.[0,1,4,4,5,6,7] if it was rotated 7 times.Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
Given the sorted rotated array nums that may contain duplicates, return the minimum element of this array.
You must decrease the overall operation steps as much as possible.
\n\n\n
Example 1:
\nInput: nums = [1,3,5]\nOutput: 1\n
Example 2:
\nInput: nums = [2,2,2,0,1]\nOutput: 0\n\n
\n
Constraints:
\n\nn == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000nums is sorted and rotated between 1 and n times.\n
Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?
\n", "translatedTitle": "寻找旋转排序数组中的最小值 II", "translatedContent": "已知一个长度为
n 的数组，预先按照升序排列，经由 1 到 n 次 旋转 后，得到输入数组。例如，原数组 nums = [0,1,4,4,5,6,7] 在变化后可能得到：\n4 次，则可以得到 [4,5,6,7,0,1,4]7 次，则可以得到 [0,1,4,4,5,6,7]注意，数组 [a[0], a[1], a[2], ..., a[n-1]] 旋转一次 的结果为数组 [a[n-1], a[0], a[1], a[2], ..., a[n-2]] 。
给你一个可能存在 重复 元素值的数组 nums ，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 最小元素 。
你必须尽可能减少整个过程的操作步骤。
\n\n\n\n
示例 1：
\n\n\n输入：nums = [1,3,5]\n输出：1\n\n\n
示例 2：
\n\n\n输入：nums = [2,2,2,0,1]\n输出：0\n\n\n
\n\n
提示：
\n\nn == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000nums 原来是一个升序排序的数组，并进行了 1 至 n 次旋转\n\n
进阶：这道题与 寻找旋转排序数组中的最小值 类似，但 nums 可能包含重复元素。允许重复会影响算法的时间复杂度吗？会如何影响，为什么？
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