Suppose an array sorted in ascending order 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.

The array may contain duplicates.

Example 1:

Input: [1,3,5]
Output: 1

Example 2:

Input: [2,2,2,0,1]
Output: 0

Note:

• This is a follow up problem to Find Minimum in Rotated Sorted Array.
• Would allow duplicates affect the run-time complexity? How and why?

题源：here；完整实现：here

思路：

出现了重复数字会影响我们程序的走向，为了继续使用序列的顺序性，我们可将重复的数字删除一部分，当然这样的话最坏情况时间复杂度会变为O(n)。

1 暴力求解

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	int findMin(vector<int>& nums) {
		int res = INT_MAX;
		for (auto i : nums) res = min(res, i);
		return res;
	}

2 二分法

	int findMin2(vector<int> &nums){
		if (nums.size() == 1) return nums[0];
		if (nums.size() == 2) return min(nums[0], nums[1]);

		decltype(nums.size()) left = 0;
		decltype(nums.size()) right = nums.size() - 1;
		decltype(nums.size()) mid = (left + right) / 2;
		while (right - left >= 2){
			while (right - left >= 2 && nums[left] == nums[right]) right--;
			if (right - left < 2) break;
			mid = (left + right) / 2;
			if (nums[mid - 1] > nums[mid] && nums[mid + 1] > nums[mid]) return nums[mid];
			if (nums[mid] >= nums[left] && nums[left] >= nums[right]){
				left = mid;
				mid = (left + right) / 2;
			}
			else{
				right = mid;
				mid = (left + right) / 2;
			}
		}
		return min(nums[left], nums[right]);
	}