搜索旋转有序数组

描述
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.

代码

package binarySearch;
// 时间复杂度 O(log n)，空间复杂度 O(1)
public class SearchRotated { public static void main(String[] args) throws Exception {
    int[] A = {11,12,13,15,17,22,28,30,2,4,5,6,9};
    System.out.println(new SearchRotated().search(A, 17));
}

    public int search(int[] A, int target) {
        if(A==null || A.length==0)
            return -1;
        int l = 0;
        int r = A.length-1;
        while(l<=r)
        {
            int m = (l+r)/2;
            if(target == A[m])
                return m;
            if(A[m]<A[r])
            {
                if(target>A[m] && target<=A[r])
                    l = m+1;
                else
                    r = m-1;
            }
            else
            {
                if(target>=A[l] && target<A[m])
                    r = m-1;
                else
                    l = m+1;
            }
        }
        return (A[l] == target) ? l : -1;
    }

}