154. Find Minimum in Rotated Sor

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 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.

public class Solution {
    public int findMin(int[] nums) {
        
        int start = 0,end = nums.length-1;
        while(nums[start]>=nums[end])   //  注意限制条件！！！
        {
            int mid = (start + end)>>1;
            System.out.println(nums[mid]);
            System.out.println(nums[start]);
            if(nums[mid]==nums[start]&&nums[mid]==nums[end])
            {
                int min = nums[0];
                for(int i=1;i<nums.length;i++)
                {
                    if(nums[i]<min)
                       min = nums[i];
                }
                return min;
            }
            if(nums[mid]>=nums[start])
               {
                   start = mid +1;
                   System.out.println(start);
               }
            else 
               {
                   end = mid;
              //     System.out.println(end);
               }
        }
        return nums[start];//注意这个神奇的start或者left!!!
    }
}