leetcode 81. Search in Rotated Sorted Array II

Follow up for "Search 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).

Write a function to determine if a given target is in the array.

The array may contain duplicates.

有重复元素的话对这道题的影响就是没办法迅速确定从哪个位置翻转的。。。也把我搞蒙了

二分法解决这道问题，转别人做法

public boolean search(int[] nums, int target) {
        int start = 0, end = nums.length - 1, mid = -1;
        while(start <= end) {
            mid = (start + end) / 2;
            if (nums[mid] == target) {
                return true;
            }
            //If we know for sure right side is sorted or left side is unsorted
            if (nums[mid] < nums[end] || nums[mid] < nums[start]) {
                if (target > nums[mid] && target <= nums[end]) {
                    start = mid + 1;
                } else {
                    end = mid - 1;
                }
            //If we know for sure left side is sorted or right side is unsorted
            } else if (nums[mid] > nums[start] || nums[mid] > nums[end]) {
                if (target < nums[mid] && target >= nums[start]) {
                    end = mid - 1;
                } else {
                    start = mid + 1;
                }
            //If we get here, that means nums[start] == nums[mid] == nums[end], then shifting out
            //any of the two sides won't change the result but can help remove duplicate from
            //consideration, here we just use end-- but left++ works too
            } else {
                end--;
            }
        }
        
        return false;
    }

end-- 和 start++ 这里可以根据相等的情况细分优化一下