Leetcode Find Minimum in Rotated Sorted Array II

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.

这道题和上道题不同之处在于这道题中的元素可以重复，那么如果仍然用上道题的解法就可能出现问题，即序列 1,0,1,1,1 和1,1,1,0，1这两个序列

num[lo]=num[mid]=num[hi] 无法判定mid是位于前面的递增数组中还是位于后面那个递增数组中，此时的情况就只能遍历从lo-hi的元素，查找最小的那个元素了

package Find.Minimum.in.Rotated.Sorted.Array.II;

/**
 * @author dell
 *这道题说题目中可能出现重复元素，因此不能像上道题那样解
 */
public class FindMinimuminRotatedSortedArrayII {
public int findMin(int[] num) {
     if(num.length==1) return num[0];
       int lo=0;
       int hi=num.length-1;
       int mid=lo;
       while(num[lo]>=num[hi]){
           if(hi-lo==1){
               mid=hi;
               break;
           }
           mid=(lo+hi)/2;
           if(num[mid]==num[lo]&&num[mid]==num[hi]){
               mid=this.minInOrder(num, lo, hi);
               break;
           }
           if(num[mid]>=num[lo]) lo=mid;
           else hi=mid;
       }
       return num[mid];
    }
private int minInOrder(int[] num,int lo,int hi){
    int result=num[lo];
    int index=lo;
    for(int i=lo+1;i<=hi;i++){
        if(result>num[i]){
            result=num[i];   
            index=i;
        }
    }
    return index;
}
public static void main(String args[]){
    FindMinimuminRotatedSortedArrayII service=new FindMinimuminRotatedSortedArrayII();
    //int num[]={-1,-1,-1,-1};
    int num[]={3,3,1};
    //int num[]={1,1,3,1};
    int a=service.findMin(num);
    System.out.println(a);
}
}