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.

class Solution {
public:
    int search(int A[], int n, int target) {
        if(n == 0)
            return -1;
        if(n == 1) {
            if(A[0] == target)
                return 0;
            else
                return -1;
        }
        
        int median = n / 2;
        if(A[median] == target)
            return median;
            
        if(A[0] < A[median] && target >= A[0] && target < A[median] ||
            (A[0] > A[median] && (target >= A[0] || target < A[median]))){
            return search(A, median, target);
        }else{
            if(median + 1 >= n)
                return -1;
            else{
                int ret =  search(A + median + 1, n - median -1, target);
                if(ret == -1)
                    return -1;
                else 
                    return median + 1 + ret;
            }
        }
    }
};