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.

Summary: Almost the same to Search in Rotated Sorted Array without duplicates, but careful about the corner case like [1311], we should search both left half and right half. So, in this case, the run-time complexity will between O(logN) to O(N).

 1 class Solution {

 2 public:

 3     bool search(int A[], int n, int target) {

 4         if(n == 0)

 5             return false;

 6         if(n == 1) {

 7             if(A[0] == target )

 8                 return true;

 9             else 

10                 return false;

11         }

12         

13         int median = n/2;

14         if(A[median] == target)

15             return true;

16         if(A[0] == target)

17             return true;

18             

19         if(A[0] < A[median] && target > A[0] && target < A[median] || 

20             (A[0] > A[median] && (target > A[0] || target < A[median]))){

21               return search(A + 1, median - 0, target);  

22             }else if(A[0] == A[median]){

23                 bool ret1 = search(A + 1, median - 0, target);      

24                 bool ret2 = false;

25                 if(n - median - 1 <= 0)

26                     ret2 = false;

27                 

28                 ret2 = search(A + median + 1, n - median -1, target);

29                 return ret1 || ret2;

30             }else{

31                 if(n - median - 1 <= 0)

32                     return false;

33                 

34                 return search(A + median + 1, n - median -1, target);

35             }

36     }

37 };