Search in Rotated Sorted Array II leetcode

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.

假设原数组是{1,2,3,3,3,3,3}，那么旋转之后有可能是{3,3,3,3,3,1,2}，或者{3,1,2,3,3,3,3}，这样的我们判断左边缘和中心的时候都是3，如果我们要寻找1或者2，我们并不知道应该跳向哪一半。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[left] && target < nums[mid]) {
                    right = mid - 1;
                } else {
                    left = mid + 1;
                }
            } else if (nums[mid] < nums[left]) {
                if (target <= nums[right] && target > nums[mid]) {
                    left = mid + 1;
                } else {
                    right = mid - 1;
                }
            } else {
                left++;
            }
        }
        return false;
    }
}

public class Solution {

    public boolean search(int[] A, int target) {
         if (A.length == 0 || A == null){
            return false;
        }
        int start = 0;
        int end = A.length - 1;
        while (start + 1 < end){
            int mid = start + (end - start) / 2;
            // if (A[mid] == target){
            //     return true;
            // }
            if (A[start] == A[mid]){//if start == mid 可能出现mid在第一区也可能第二区
                start++;// 所以用start++ 来判断
            } else if (A[mid] > A[start]){// case Mid1
                if (target >= A[start] && target <= A[mid]){//
                    end = mid;
                } else {
                    start = mid;
                }
            } else if (A[mid] < A[start]) { //case Mid2
                if (A[start] > target && A[mid] < target){
                    start = mid;
                } else {
                    end = mid;
                }
            } else {
                end--;
            }
        }
        if (A[start] == target){
            return true;
        } else if (A[end] == target){
            return true;
        } else {
            return false;
        }
    }
}