Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

Find the minimum element.

You may assume no duplicate exists in the array.

使用二分搜索，设当前搜索范围为[left, right], 则中间的值为mid = (left + right) /2, 则

• 若a[left] < a[right], 说明没有旋转，返回a[left].
• 若a[mid] > a[right], 则结果在右边， left = mid + 1
• 若a[mid] < a[right], 则结果在左边， right = mid

关键注意边界为题, 什么时候用>=什么时候用>, 以及什么时候用mid, 什么时候用mid + 1

int findMin(vector<int> &nums) {
	int n = nums.size();
	if (n == 0)
		return INT_MIN;
	int s = 0, t = n - 1;
	while (s < t) {
		if (nums[s] < nums[t])
			return nums[s];
		int mid = (s + t) >> 1;
		if (nums[mid] > nums[t])
			s = mid + 1;
		else 
			t = mid;
	}
	return nums[s];
}

1. 当有重复元素存在时，见Find Minimum in Rotated Sorted Array II

2. Search in Rotated Sorted Array, 在旋转列表中查找某个数