is there any better way to solve this problem?

Question

Write a Java program to rearrange all the elements of an given array of integers so that all the odd numbers come before all the even numbers.

import java.util.Arrays;

public class Arr_arrange {

public static void main(String[] args) {
int[] arr = {12,76,89,54,23,93,19};
int[] na = new int[arr.length];
int j = 0;
for (int k : arr) {
if (k % 2 != 0) {
na[j] = k;
j++;
}
}

for (int k : arr) {
if (k % 2 == 0) {
na[j] = k;
j++;
}
}
System.out.println("new arrays is : " + Arrays.toString(na));

}
}

OUTPUT : new arrays is : [89, 23, 93, 19, 12, 76, 54]

Answer 1

Not better, just another way into doing it.

https://code.sololearn.com/cWpPQznnL49H/?ref=app

Answer 2

Using the Comparator functional interface.

https://code.sololearn.com/cVvhWS5mRX9I/?ref=app

Sadly the array must be boxed for this to work (won't run on primitive int) so it's a little tricky.
If it's an array of objects, the comparator can be also passed directly to Arrays.sort() without creating a separate stream.

Answer 3

Like Tibor Santa, using stream and comparator but a bit more condensed 😉

https://code.sololearn.com/cVW1GLhWpaLe/?ref=app

Answer 4

public static int[] sortOddFirst( int[] a) {
int iOdd=0, iEven = a.length-1;
int[] aNew = new int[a.length];

for (int n: a)
if ( (n & 1) == 0)
aNew[iEven--] = n;
else
aNew[iOdd++ ] = n;

return aNew;
}