Find Minimum in Rotated Sorted Array II

Description:

Follow up for "Find Minimum in Rotated Sorted Array":
What if duplicates are allowed?

Would this affect the run-time complexity? How and why?

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.

The array may contain duplicates.

Code:

 1  int findMinElement(vector<int>& nums, int indexLeft, int indexRight)

 2     {

 3         if (indexLeft < indexRight)

 4         {

 5             if ( nums[(indexLeft+indexRight)/2] < nums[indexRight])

 6                 return findMinElement(nums, indexLeft, (indexLeft+indexRight)/2);

 7             else if ( nums[(indexLeft+indexRight)/2] > nums[indexRight])

 8                  return findMinElement(nums, (indexLeft+indexRight)/2+1, indexRight);

 9              else

10             {//此时不能判最小值在左边还右边，遍历判断

11                 int result = nums[indexLeft];

12                 for (int i = indexLeft+1; i <= indexRight; i++)

13                 {

14                     if ( nums[i] < result)

15                          result = nums[i];

16                 }

17                 return result;

18             }

19         }

20         else

21             return nums[indexLeft];

22     }

23     int findMin(vector<int>& nums) {

24         return findMinElement(nums, 0, nums.size()-1 );

25     }