Split an array into two arrays

Question

I need elements of matrix put into an array, then i need to sort first odd numbers and then even numbers Example: This is array: 5, 9, 1, 2, 3, 8, 4. output: 1,3,5,9 ; 2,4,8

This is my code:

int[] array=new int[mat.length*mat[0].length];
int cnt=0;

for(int i=0; i<mat.length; i++)
{
  for(int j=0; j<mat[0].length; j++)
  {
    array[cnt]=mat[i][j];
    cnt++;
  }
}
int cnt1=0;
int cnt2=0;
int[] array1=new int[array.length];
int[] array2=new int[array.length];
for(int i=0; i<array.length; i++)
{
  if(array[i]%2==0)
  {
    array1[br1]=array[i];
    cnt1++;
  }
  else
  {
    array2[br2]=array[i];
    cnt2++;
  }
}

The problem is this two arrays for odd and even numbers, because i don't know their length and if i put the size of whole array, then i will get zeros for remaining places in odd array for number that is even and vice versa. How would you do this? Thank you

You can keep a running count of the number of even elements (or odd elements, or both) in the double-loop. Watch out for the typo in the loop over j. It should be j<mat[i] — fresidue
– fresidue, Commented Jun 29, 2015 at 22:08

Omar Mainegra · Accepted Answer · 2015-06-29 22:19:32Z

4

If you can use List you can do

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

for(int i=0; i<mat.length; i++) {
    for(int j=0; j<mat[0].length; j++) {
        if (mat[i][j] % 2 == 0)
            even.add(mat[i][j]);
        else
            odd.add(mat[i][j]);
    }
}

Collections.sort(even);
Collections.sort(odd);

odd.addAll(even);

for (int v: odd){
    System.out.println(v);
}

edited Jun 29, 2015 at 22:19

answered Jun 29, 2015 at 22:12

Omar Mainegra

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Comments

Makoto · Accepted Answer · 2015-06-29 22:18:00Z

Here are a couple of Java 8 solutions (which are far more straightforward):

With two passes of the stream, it's filter and sort.

final int[] ints = {5, 9, 1, 2, 3, 8, 4};

int[] oddArray = Arrays.stream(ints).filter(x -> x % 2 != 0).sorted().toArray();
int[] evenArray = Arrays.stream(ints).filter(x -> x % 2 == 0).sorted().toArray();

System.out.println(Arrays.toString(oddArray));
System.out.println(Arrays.toString(evenArray));

With one pass of the stream, you'd want to use collections so that you don't have to deal with proper sizing of the array. You still have to sort it, though.

final int[] ints = {5, 9, 1, 2, 3, 8, 4};

List<Integer> oddList = new ArrayList<>();
List<Integer> evenList = new ArrayList<>();

Arrays.stream(ints).forEach(e -> {
    if(e % 2 != 0) {
        oddList.add(e);
    } else {
        evenList.add(e);
    }
});
Collections.sort(oddList);
Collections.sort(evenList);

System.out.println(oddList);
System.out.println(evenList);

Alp · Accepted Answer · 2015-06-29 22:11:38Z

1

 // This is how you instantiate collections.
 List odds = new ArrayList(); 
 List evens = new ArrayList();

 ...
 if(array[i]%2==0)
  {
    // Here you add a new item to the collection for even numbers
    evens.add(array[i];
  }
  else
  {
    // Here you add a new item to the collection for odd numbers
    odds.add(array[i]);
  }


...

// And finally this is how you get arrays out of collections
int[] oddArray = odds.toArray(new int[]);
int[] evenArray = evens.toArray(new int[]);

answered Jun 29, 2015 at 22:11

Alp

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1 Comment

Makoto Over a year ago

Notwithstanding the unclosed paren, you're using raw types which is a cardinal sin in the world of generics.

MaxZoom · Accepted Answer · 2015-06-29 22:24:09Z

1

There is not problem with array size as there is enough space in each array (array1 and array2) to hold all the numbers and you know the count (cnt1 and cnt2) of elements in each array. Thus after loop you can copy only valid elements to a new array as below:

int[] even = Arrays.copyOf(array1, cnt1);
int[] odd = Arrays.copyOf(array2, cnt2);

Arrays.copyof(..) reference

edited Jun 29, 2015 at 22:24

answered Jun 29, 2015 at 22:11

MaxZoom

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Comments

zeus · Accepted Answer · 2015-06-29 22:20:27Z

0

Thank you all

member @fresidue remind me to this which help me to solve this problem

int cnt=0;
    int cnt1=0;
    int cnt2=0;
    for(int i=0; i<mat.length; i++)
    {
      for(int j=0; j<mat[0].length; j++)
      {
        array[cnt]=mat[i][j];
        cnt++;
        if(mat[i][j]%2==0)
          cnt1++;
        else
          cnt2++;
      }
    }

    int[] array1=new int[cnt1];
    int[] array2=new int[cnt2];

answered Jun 29, 2015 at 22:20

zeus

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Comments

Anton Steenvoorden · Accepted Answer · 2015-06-29 22:09:20Z

-1

You could also use an ArrayList, this has a dynamic size but is a bit slower. This solves the problem of the zeros

answered Jun 29, 2015 at 22:09

Anton Steenvoorden

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