Java sorting program

Question

This is my first post so please excuse me if I've made any errors.

In this program I've written, the user inputs integers one at a time from the keyboard, 2 4 5 1 3 for example. The reverse method if called before any of the others would return 3 1 5 4 2. However, if the sort method is called, giving us 1 2 3 4 5, then the reverse method is called, we get 5 4 3 2 1. Any ideas how to get the reverse to return the reverse order of the original input even after the other methods have been called?

   public static void reverse(ArrayList<Integer> num) {
        ArrayList<Integer> newNum= new ArrayList<Integer>();
        newNum = num;   

        Collections.reverse(newNum);
        System.out.println(newNum); 
    }

Joe F · Accepted Answer · 2013-04-26 01:27:54Z

2

Every time you do this:

newNum = num;

You are making newNum point to the same ArrayList as num. Any changes you make to either newNum or num will be reflected in the other. They are the same ArrayList instance.

You probably want to construct a new ArrayList instance containing the same elements.

ArrayList<Integer> newNum = new ArrayList<Integer>(num);

answered Apr 26, 2013 at 1:27

Joe F

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1 Comment

Chase S Over a year ago

Thanks much, I see you noticed my failed attempt at creating new sets of data to be modified rather than messing with the original. This worked for me. Thanks again.

Μenelaοs · Accepted Answer · 2013-04-26 01:27:39Z

1

Why not have two arraylists... One with the result of the last operation, and one with the original set as given. Therefore, result will always change and the original will stay the same.

answered Apr 26, 2013 at 1:27

Μenelaοs

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1 Comment

Chase S Over a year ago

Creating new ArrayLists was the ticket! Thanks!

Lokathor · Accepted Answer · 2013-04-26 01:27:56Z

1

You have to make a copy of what the user input before you start messing with it if you want to save it for later.

public static void main(String[] args) {

ArrayList<Integer> num = new ArrayList<Integer>();
ArrayList<Integer> numStartingList; 

//stuff about reading the variable in here

numStartingList = new ArrayList<>(num); 

// read options and show output
}

answered Apr 26, 2013 at 1:27

Lokathor

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Comments

Michael W · Accepted Answer · 2013-04-26 01:37:47Z

0

In each of your methods when you call newNum = num, you are giving up the reference to the array list you just "newed" and pointing newNum at the memory location of "num" (or doing what is called a "shallow copy." Anything that you do to newNum is being done to num, and vice versa.

Instead you want to make a "deep copy" of your num array. I'd suggest replacing the first two lines of each method with a call to a private method doing something like this:

private ArrayList<Integer> copyList(ArrayList<Integer> in) { 
    ArrayList<Integer> out = new ArrayList<Integer>();
    for (int i : in) {
        out.add(i);
    }
    return out;
}

Though I haven't checked that for correctness.

Good luck!

answered Apr 26, 2013 at 1:37

Michael W

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2 Comments

Chase S Over a year ago

Thanks for the reply. I seem to have solved the problem by using the above suggestion and creating a new ArrayList instance and calling my methods on that. I will keep "deep" and "shallow" copies in mind for future reference, those are new terms for me.

Romski Over a year ago

I think the use of deep/shallow copy is inappropriate here. These are used to describe the process of cloning an object, where a deep copy would traverse the object graph and all its attributes to produce a copy of each part.

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