How to sort an array without destroying the original array?

Question

I have original array of

public static void main (String[] arg) {
  int[] array = {1,5,6,8,4,2}

  for (int i = 0; i < array.length; i++) {
    System.out.print("List 1 = " + array[i] + ",");
  }
  swap(array);
  for (int i = 0; i < array.length; i++) {
    System.out.print("List 2 = "+array[i] + ",");
  }
}
private static int swap (int[] list){
   Arrays.sort(list);
}

The output is

 List 1 = 1,5,6,8,4,2
 List 2 = 1,2,4,5,6,8

What I want the answer to be is

List 1 = 1,5,6,8,4,2
List 2 = 1,5,6,8,4,2

even after sorting. How can I do it?

But if shows the same after sorting is not sorting at all... — DGomez
– DGomez, Commented Apr 19, 2013 at 16:10

Juvanis · Accepted Answer · 2013-04-19 18:16:13Z

20

int[] originalArray = {1,5,6,8,4,2};
int[] backup = Arrays.copyOf(originalArray,originalArray.length);
Arrays.sort(backup);

After the execution of the code above, backup becomes sorted and originalArray stays same.

edited Apr 19, 2013 at 18:16

answered Apr 19, 2013 at 16:06

Juvanis

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4 Comments

userpane Over a year ago

IF so, How can i pass that sorted array to another method?

Juvanis Over a year ago

@userpane are you kidding? you can just call something like yourMethod(backup);

userpane Over a year ago

No no... I know that but It is in non-main method so I could not pass it to another non-main method.

Juvanis Over a year ago

you can return it. example: public int[] yourMethod() { //some other code return backup; }

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