This program finds the number in an array that's closest to zero.

FindTheNumberClosestToZero

import org.junit.Test;

public class BetterProgrammer05122012Test extends junit.framework.TestCase {
    public static int[] numbers = new int[] {5, 10, -5, 0, 25, 35};
    public static int[] numbers2 = new int[] {5, 10, -5, -2, 0, 25, 35};
    public static int[] noNegativeNumbers = new int[] {5, 10, 15, 25, 35};

    @Test
    public void testBetterProgrammer05122012 () {

            assertEquals(5, BetterProgrammer05122012.getSumOfTwoClosestToZeroElements(numbers));
    }

    @Test
    public void testArrayWithNegativeNumberLowest () {
        assertEquals(2, BetterProgrammer05122012.getSumOfTwoClosestToZeroElements(numbers2));
    }

    @Test
    public void testArrayWithNoNegativeNumbers() {
        assertEquals(5, BetterProgrammer05122012.getSumOfTwoClosestToZeroElements(noNegativeNumbers));
    }

}






import java.util.Arrays;

public class BetterProgrammer05122012 {

        public static int getSumOfTwoClosestToZeroElements(int[] a) {
            /*
             Please implement this method to
             return the sum of the two elements closest to zero.
             If there are two elements equally close to zero like -2 and 2,
             consider the positive element to be "closer" to zero than the negative one.
            */
            int[] b = a;
            Arrays.sort(b);
            int negativeValuePlacement = 0;
            int positiveValuePlacement = 0;
            int closestNumberToZero;

            for( int i = 0; i < b.length; i++) {
                if (b[i] < 0) {negativeValuePlacement++;
                    positiveValuePlacement++;
                }
                if (b[i] == 0) positiveValuePlacement++;
            }
            --negativeValuePlacement;

            if (b[0] >= 0) closestNumberToZero = b[positiveValuePlacement];
            else {
                if ((b[negativeValuePlacement]*-1) < b[positiveValuePlacement]) {
                closestNumberToZero = (b[negativeValuePlacement]*-1);
            }   else    closestNumberToZero = b[positiveValuePlacement]; }
        return closestNumberToZero;
        }
    }