[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.

public class Solution {public int findMin(int[] num) {if(num.length==0){return Integer.MAX_VALUE;}int left = 0,right = num.length-1;while(left<right){if(num[left]<num[right]) return num[left];int mid = (left+right)/2;if(num[mid]>num[right]){if(mid==left) left = mid++;left = mid;}else if(num[mid]==num[right]){if(num[mid]==num[left]){return Math.min(findMin(Arrays.copyOfRange(num, left, mid)), findMin(Arrays.copyOfRange(num, mid+1,right+1)));}else if(num[mid]>num[left]){return num[left];}else {right = mid;}}else{right = mid;}}return num[left];}}

solution2

public class Solution {public int findMin(int[] num) {assert num.length>0;int left = 0,right = num.length-1;while(left<right&&num[left]>=num[right]){int mid = (left+right)/2;if(num[mid]>num[right]){left = mid+1;}else if(num[mid]<num[right]){right = mid;}else{//left++;}}return num[left];}}