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.

Solution: binary search

解决的办法只能是对边缘移动一步，直到边缘和中间不在相等或者相遇，这就导致了会有不能切去一半的可能。所以最坏情况（比如全部都是一个元素，或者只有一个元素不同于其他元素，而他就在最后一个）就会出现每次移动一步，总共是n步，算法的时间复杂度变成O(n)。

public class Solution {
    public boolean search(int[] nums, int target) {
        if (nums == null || nums.length == 0) {
            return false;
        }
        
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            int mid = (left + right) / 2;
            if (nums[mid] == target) {
                return true;
            }
            
            if (nums[mid] > nums[left]) {
                if (target < nums[mid] && target >= nums[left]) {
                    right = mid - 1;
                } else {
                    left = mid;
                } 
            } else if (nums[mid] < nums[left]) {
                if (target > nums[mid] && target <= nums[right]) {
                    left = mid + 1;
                } else {
                    right = mid;
                }
            } else {
                left++;  //处理中间和边缘相等情况，移动一步
            }
        }
        
        return false;
    }
}