Search in Rotated Sorted Array [LeetCode]

Suppose a sorted array is rotated at some pivot unknown to you beforehand.

(i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Summary: Binary search can apply to a rotated sorted array, just by analysis of its property.

 1 class Solution {
 2 public:
 3     int search(int A[], int n, int target) {
 4         if(n == 0)
 5             return -1;
 6         if(n == 1) {
 7             if(A[0] == target)
 8                 return 0;
 9             else
10                 return -1;
11         }
12         
13         int median = n / 2;
14         if(A[median] == target)
15             return median;
16             
17         if(A[0] < A[median] && target >= A[0] && target < A[median] ||
18             (A[0] > A[median] && (target >= A[0] || target < A[median]))){
19             return search(A, median, target);
20         }else{
21             if(median + 1 >= n)
22                 return -1;
23             else{
24                 int ret =  search(A + median + 1, n - median -1, target);
25                 if(ret == -1)
26                     return -1;
27                 else 
28                     return median + 1 + ret;
29             }
30         }
31     }
32 };