RecursiveBinarySearch

/**
   An methods for a recursive binary search. Searches for both halves of an array and returns the larger of the two.
 */
public class RecursiveBinarySeach
{
   /** Searches for a specified entry in an array.
    *  @param anArray  The array to be searched.
    *  @param first    First entry in the array.
    *  @param last     Last entry in the array.
    *  @param index    Position inside the array while searching.
    *  @param largest1 The largest element in the first half of the array.
    *  @param largest2 The largest element in the second half of the array.
    *  @return  Returns the largest element in the array. This can be either largest1 or largest2
                depending on which of the two is larger.
    */
   private static <T extends Comparable<? super T>> T binarySearch(T[] anArray, int first, int last, int index, T largest1, T largest2)
   {
      int mid = first + (last - first) / 2;

      if (index < mid)
      {
         if (largest1.compareTo(anArray[index]) < 0)
         {
            largest1 = anArray[index];
         }
         return binarySearch(anArray, first, last, index++, largest1, largest2);
      }
      else if (index >= mid && index <= anArray.length)
      {
         if (largest2.compareTo(anArray[index]) < 0)
         {
            largest1 = anArray[index];
         }
         return binarySearch(anArray, first, last, index++, largest1, largest2);
      }
      else
      {
         if (largest1.compareTo(largest2) > 0)
         {
            return largest1;
         }
         else
         {
            return largest2;
         }
      }
   }

   /** Helper method for binarySearch().
    *  @param anArray  The array to be searched.
    *  @return  Returns the largest element in the array. This can be either largest1 or largest2
                depending on which of the two is larger.
    */
   public static <T extends Comparable<? super T>> T binarySearch(T[] anArray)
   {
      return binarySearch(anArray, 0, anArray.length - 1, 0, anArray[0], anArray[anArray.length - 1]);
   }
}