Adding even and odd integers into Array

Question

I'm trying to figure out how to sort a list of input in one array, and make two different arrays out of that, either into even or odd numbers. I can't seem to add the integers to the array in the if-loop.

Here is my code:

 Scanner in = new Scanner (System.in);

System.out.print("Enter a number");
    int number = in.nextInt();


 int [] x = new int[0];
 int [] even = new int [0];
 int [] odd = new int [0];



for (int i = 0; i <x.length; i++)
{

    if (in.nextInt() == 0){
        for (i = 0; i <x.length; i++)
        {
            if (x[i] % 2 == 0)
            {
            even = even + x[i];
            System.out.print("Even numbers = " + even);
            i++;
            }
            if (x[i] % 2 != 0)
            {
            odd = odd + x[i];
            System.out.print("Odd numbers = " + odd);
            i++;
            }
        }
        break;
                }

    else {
        x[i] = number;
        i++;
        }

}

John Kugelman · Accepted Answer · 2013-03-27 21:46:07Z

2

Arrays are fixed size in Java. They don't grow dynamically. Use an ArrayList if you want an array-like container that can grow and shrink.

List<Integer> even = new ArrayList<Integer>();
List<Integer> odd  = new ArrayList<Integer>();

if (...)
{
    even.add(number);
}
else
{
    odd.add(number);
}

answered Mar 27, 2013 at 21:46

John Kugelman

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Comments

pamphlet · Accepted Answer · 2013-03-27 21:47:28Z

2

You aren't using arrays correctly. I'm assuming this is homework designed to teach you how to use them. Try this tutorial:

http://docs.oracle.com/javase/tutorial/java/nutsandbolts/arrays.html

Good luck!

answered Mar 27, 2013 at 21:47

pamphlet

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Comments

DntFrgtDSemiCln · Accepted Answer · 2013-03-27 21:52:57Z

0

I would not use three arrays to do this. Simply traverse through the list and have two loop variables that reference the even and odd indexes.Something like this:

int evenIndex=0,oddIndex=1;
int evenSum=0,OddSum=0;
int [] x=new int[10];

 while(evenIndex <= x.length && oddIndex <= x.length)
        {
          evenSum+=x[evenIndex];
           oddSum+=x[oddIndex];


             evenIndex+=2;
             oddIndex+=2 

            }

answered Mar 27, 2013 at 21:52

DntFrgtDSemiCln

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2 Comments

AJMansfield Over a year ago

This is not what the OP is asking for. The OP wants to seperate the elements that are themselves even or odd, not add the elements at even and odd indicies. Plus, I'm not 100% sure, but your solution will always throw an ArrayIndexOutOfBoundsException at either the 7th or 8th line, depending on if the original array has an even or odd number of elements.

DntFrgtDSemiCln Over a year ago

I thought that even = even + x[i]; and odd = odd + x[i]; in the code suggests that the OP wants to sum them up ? Anyway my bad if it did'nt serve the purpose.

AJMansfield · Accepted Answer · 2013-03-28 15:28:35Z

0

Two good solutions:

int [] x = new int[0];
LinkedList<Integer> even = new LinkedList<Integer>(),
                     odd = new LinkedList<Integer>();

for(int num: x)
    if (x & 1 == 1)
        odd.add(num);
    else
        even.add(num);

The other option is to iterate through, counting how many evens and odds there are, allocating arrays of the correct size, and then putting the numbers into the arrays. That is also an O(n) solution.

edited Mar 28, 2013 at 15:28

answered Mar 27, 2013 at 21:53

AJMansfield

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4 Comments

AJMansfield Over a year ago

Also, if you need the results as arrays, just use even.toArray() or the same for odd.

John Kugelman Over a year ago

An ArrayList would not be O(n^2). Adding an item to an ArrayList takes O(1) amortized time, so adding n elements is O(n) amortized.

AJMansfield Over a year ago

@JohnKugelman Actually you are right that it isn't O(n^2), come to think of it. ArrayLists grow themselves by doubling, not incrementing, so the average case time for the way you have it is actually O(n log n), I think, because add actually takes O(n) for log(n) of the n times we need to add something. Thus, each add really takes O(log n). Doing that n times is clearly then O(n log n). However, if you give the ArrayLists initial capacity parameters that ensure they will never have to be resized, you can do it in O(n) average case, although at that point you aren't saving any extra RAM.

John Kugelman Over a year ago

@AJMansfield If n is the total number of items, add doesn't take O(n) when it has to do a resize. It takes O(k) where k is the size of the array at that time. With a capacity-doubling implementation k is a series of doubling numbers, the sum of which is approximately 2n. That means these intermediate resizes take O(2n). Of course mathematically speaking O(2n) = O(n), so there we are: adding n elements to an ArrayList takes O(n) amortized time. "Amortized" accounts for the occasional resize; if you average them all together you still end up with O(n).

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