Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
[4,5,6,7,0,1,2] if it was rotated 4 times.[0,1,2,4,5,6,7] 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 of unique elements, return the minimum element of this array.
You must write an algorithm that runs in O(log n) time.
\n
Example 1:
\n\n\nInput: nums = [3,4,5,1,2]\nOutput: 1\nExplanation: The original array was [1,2,3,4,5] rotated 3 times.\n\n\n
Example 2:
\n\n\nInput: nums = [4,5,6,7,0,1,2]\nOutput: 0\nExplanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.\n\n\n
Example 3:
\n\n\nInput: nums = [11,13,15,17]\nOutput: 11\nExplanation: The original array was [11,13,15,17] and it was rotated 4 times. \n\n\n
\n
Constraints:
\n\nn == nums.length1 <= n <= 5000-5000 <= nums[i] <= 5000nums are unique.nums is sorted and rotated between 1 and n times.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 = [0,1,2,4,5,6,7] 在变化后可能得到：\n4 次，则可以得到 [4,5,6,7,0,1,2]7 次，则可以得到 [0,1,2,4,5,6,7]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]] 。
给你一个元素值 互不相同 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 。
你必须设计一个时间复杂度为 O(log n) 的算法解决此问题。
\n\n
示例 1：
\n\n\n输入：nums = [3,4,5,1,2]\n输出：1\n解释：原数组为 [1,2,3,4,5] ，旋转 3 次得到输入数组。\n\n\n
示例 2：
\n\n\n输入：nums = [4,5,6,7,0,1,2]\n输出：0\n解释：原数组为 [0,1,2,4,5,6,7] ，旋转 4 次得到输入数组。\n\n\n
示例 3：
\n\n\n输入：nums = [11,13,15,17]\n输出：11\n解释：原数组为 [11,13,15,17] ，旋转 4 次得到输入数组。\n\n\n
\n\n
提示：
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