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);
}
}

}

}