SearchRotatedSortedArray.java

package binary_search;

/**
 * Created by gouthamvidyapradhan on 10/04/2017.
 Suppose an array sorted in ascending order 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.

 */
public class SearchRotatedSortedArray
{
    /**
     * Main method
     * @param args
     * @throws Exception
     */
    public static void main(String[] args) throws Exception
    {
        int[] A = {5, 4, 3, 2, 1};
        System.out.println(new SearchRotatedSortedArray().search(A, 4));
    }

    public int search(int[] nums, int target)
    {
        if(nums.length == 0) return -1;
        if(nums.length == 1)
        {
            return (nums[0] == target) ? 0 : -1;
        }
        int low = 0, high = nums.length - 1;
        while(low < high)
        {
            int mid = (low + high) / 2;
            if(nums[mid] == target)
                return mid;
            if((nums[mid] <= nums[low]) && (target > nums[mid] && target <= nums[high]) ||
                    (nums[low] <= nums[mid] && (target < nums[low] || target > nums[mid])))
                low = mid + 1;
            else high = mid - 1;
        }
        return (nums[low] == target) ? low : -1;
    }

}