Java for LeetCode 081 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?

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

解题思路：

参考Java for LeetCode 033 Search in Rotated Sorted Array修改下代码即可，JAVA实现如下：

    public boolean search(int[] nums, int target) {
		int left = 0, right = nums.length - 1;
		while (left <= right) {
			if (target == nums[(right + left) / 2])
				return true;
			// 右半部分为旋转区域
			if (nums[(right + left) / 2] > nums[left]) {
				if (target >= nums[left] && target < nums[(right + left) / 2])
					right = (right + left) / 2 - 1;
				else
					left = (right + left) / 2 + 1;
			}
			// 左半部分为旋转区域
			else if (nums[(right + left) / 2] < nums[left]) {
				if (target > nums[(right + left) / 2] && target <= nums[right])
					left = (right + left) / 2 + 1;
				else
					right = (right + left) / 2 - 1;
			}
			// 老实遍历
			else
				left++;
		}
		return false;
    }