lintcode-medium-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.

 

public class Solution {
    /** 
     * param A : an integer ratated sorted array and duplicates are allowed
     * param target :  an integer to be search
     * return : a boolean 
     */
    public boolean search(int[] A, int target) {
        // write your code here
        
        if(A == null || A.length == 0)
            return false;
        
        int left = 0;
        int right = A.length - 1;
        
        return search(A, target, left, right);
    }
    
    public boolean search(int[] A, int target, int left, int right){
        
        if(left > right)
            return false;
        
        if(left == right){
            if(A[left] == target)
                return true;
            else
                return false;
        }
        
        int mid = left + (right - left) / 2;
        
        if(A[mid] == target)
            return true;
            
        if(A[mid] > target){
            if(A[mid] > A[left]){
                boolean flag1 = search(A, target, left, mid - 1);
                boolean flag2 = search(A, target, mid + 1, right);
                    
                if(!flag1 && !flag2)
                    return false;
                else if(!flag1)
                    return flag2;
                else
                    return flag1;
            }
            else if(A[mid] < A[left]){
                return search(A, target, left, mid - 1);
            }
            else{
                return search(A, target, left + 1, right);
            }
        }
        else{
            if(A[mid] > A[left]){
                return search(A, target, mid + 1, right);
            }
            else if(A[mid] < A[left]){
                boolean flag1 = search(A, target, left, mid - 1);
                boolean flag2 = search(A, target, mid + 1, right);
                    
                if(!flag1 && !flag2)
                    return false;
                else if(!flag1)
                    return flag2;
                else
                    return flag1;
            }
            else{
                return search(A, target, left + 1, right);
            }
        }
        
    }

}