Java array, add item into next empty index

Question

In Java, is there a way to add a value not at a specific index, but to the next undeclared index? Say:

int[] negativeArray = new int[21];
int[] positiveArray = new int[21];

There are two arrays for two different types of ints, for example negative and positive. I'm looping through this, and I want it to work something like a stack (I don't know too much about stacks, but to my knowledge you don't go through its index, you just pop/push items onto it) to where if its a negative number, put in the number in the next undeclared index spot in the negative array.

I thought of a way to do this with some extra code. I'd set all values in the array to 0. When checking if the variable is negative or positive, I would loop through the array to the next value that is 0. Once I find it I know what index I'm on. This would require a bit of effort though, are there any simpler ways of doing this?

Edit: Some comments are stating different ways to do this without using a basic array. I was assigned this, and I am required to use an array to get credit for it...

@assylias Ooh I've never used one of those before, I'll try it out. But the thing is, I am required to use an array because its a school assignment. — Gabriel
– Gabriel, Commented Feb 15, 2012 at 16:05
Posting questions about homework is fine, but please tag the question as homework:) — ARRG
– ARRG, Commented Feb 15, 2012 at 16:09

ARRG · Accepted Answer · 2012-02-15 16:07:39Z

8

If you are actually using the array as a stack (and thus, will only add or remove items at the top of the stack) then you could keep in another variable the next free index in the array.

int[] array = new int[21];
int nextIndex = 0;

public void push(int e) {
    array[nextIndex] = e;
    ++nextIndex;
}

public int pop() {
    --nextIndex;
    return array[nextIndex];
}

If removals can occur anywhere, then I don't see a better solution than iterating over the array to find a free spot.

answered Feb 15, 2012 at 16:07

ARRG

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2 Comments

Random Over a year ago

What happens to the element at [nextIndex] after you poped it? Would it leak? I think this example appears in Effective Java. Better if we add in the pop method something to nullify the element at nextIndex.

ARRG Over a year ago

@Random : You are right in the case where we are storing Objects, the object wouldn't be garbage collected if we didn't remove it from the array. This isn't a problem for integers (and other primitive types) though because the "null" (aka 0) that you would write instead takes the same amount of memory.

Guillaume Polet · Accepted Answer · 2012-02-15 16:05:37Z

3

That is why Listhave been made. Simply use something like this:

List<Integer> negativeIntegers = new ArrayList<Integer>(21);
...
negativeIntegers.add(-127);

answered Feb 15, 2012 at 16:05

Guillaume Polet

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Comments

assylias · Accepted Answer · 2012-02-15 16:28:51Z

If you want to create an array with the positive values and one with the negative values, you can do it with the algorithm you propose:

public static void main(String[] args) throws Exception {
    int[] negativeArray = new int[3];
    int[] positiveArray = new int[3];

    int[] test = new int[] {1, -1, 2, -2, 3, -3};

    int posIndex = 0;
    int negIndex = 0;

    for (int i = 0; i < test.length; i++) {
        if (test[i] > 0) {
            positiveArray[posIndex++] = test[i];
        } else if (test[i] < 0) {
            negativeArray[negIndex++] = test[i];
        }
    }

    System.out.println(Arrays.toString(test)); //[1, -1, 2, -2, 3, -3]
    System.out.println(Arrays.toString(positiveArray)); //[1, 2, 3]
    System.out.println(Arrays.toString(negativeArray)); //[-1, -2, -3]
}

user3001 · Accepted Answer · 2012-02-15 16:19:08Z

0

Check out http://docs.oracle.com/javase/6/docs/api/java/util/Stack.html

It does basically the same as ARRG's solution, but with fewer lines of code:

Stack<Integer> negative = new Stack<Integer>();
Stack<Integer> positive = new Stack<Integer>();

public void putMyNumber(int number) {
  // Taking 0 as "positive"
  if (number >= 0) {
    positive.push(number); //Auto-boxing
    return;
  }
  negative.push(number);
}

answered Feb 15, 2012 at 16:19

user3001

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