Random Number array in Java - want to sort ascending

Question

I am creating random number generator in Java as part of program I am writing to learn the language better (come from more C#/C++ background).

    ArrayList<Integer> al = new ArrayList<Integer>();
    Random ran = new Random();
    for(int i = 1; i <= 11; i++)
        al.add(i);


    for(int i = 0; i < 2; i++)
    {
        ArrayList<Integer> temp = new ArrayList<Integer>();
        int num = al.remove(ran.nextInt(al.size()));

        temp.add(num);
        Arrays.sort(temp);

        text("\Random Number " + i + " is: " + temp[i]);
    }

On arrays.sort(temp) I get a no suitable method error and in my text output function I get array required but java.util.ArrayList found. Can anyone suggest a better way to sort this random number array into ascending order or see something I am doing wrong currently that could easily be corrected? Thanks.

NPE · Accepted Answer · 2012-02-28 12:48:12Z

4

Use Collections.sort(temp).

In Java, arrays and lists are two different beasts. Arrays.sort() only works for arrays; the equivalent function for lists is Collections.sort().

I can't say I fully understand the logic behind your code, but you might also want to take a look at Collections.shuffle().

edit Upon closer inspection, there are other problems with the code:

You are re-creating temp from scratch, so on each loop iteration it will contain exactly one element.
temp[i] is not valid syntactically; the correct syntax is temp.get(i). Even with the correct syntax, it'll give you an "out of bounds" exception, since temp only contains one element.

edited Feb 28, 2012 at 12:48

answered Feb 28, 2012 at 12:36

NPE

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2 Comments

Ctrl_Alt_Defeat Over a year ago

Many Thanks - thats what you get when you search array sort in Java on google first. :) I will also take a look at collections.shuffle. With regards the output line after changing to Collections.Sort(temp); can i still print out the line with the temp[i] index?

NPE Over a year ago

@KOL: The correct syntax is temp.get(i). See the updated answer.

vdeantoni · Accepted Answer · 2012-02-28 14:17:15Z

3

When sorting a ArrayList you should use:

Collections.sort();

In your case, I suggest you to use a TreeSet, it provides you a sorted collection (it sort itself after every add or remove. Also, it prevents you from adding duplicate elements.

TreeSet<Integer> randomSet = new TreeSet<Integer>();
Random ran = new Random();
while(randomSet.size() < 3) {   
    randomSet.add(Math.abs(ran.nextInt()) % 11 + 1); //+1 to adjust your range [1..11]
}

for (int i : randomSet)
    System.out.println(i);

Math.abs() guarantee that you will only have positive numbers and you can use % operator to set a maximum value.

More info on add took from oracle docs:

If this set already contains the element, the call leaves the set unchanged and returns false.

edited Feb 28, 2012 at 14:17

answered Feb 28, 2012 at 12:37

vdeantoni

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3 Comments

Ctrl_Alt_Defeat Over a year ago

Thanks for this extra logic. I want 2 random numbers generated in the range 1-11. So from your comment the % should have 11 after it and the for loop <=3 so generate 3 numbers? The only other contraint I have is that I dont want duplictae random numbers (i.e - if 3 comes out the next time round it should not be selected.

Ctrl_Alt_Defeat Over a year ago

That was very useful - thanks - can you explain a bit more on how the % 11 + 1 is working - i.e why the need to add +1 and if I was my range was 20 I would not need the + 1 right? Is it the same as the modulo operator in C++? thanks

vdeantoni Over a year ago

Yes, that's the same in C++. Without the +1, the range goes from 0 to 10. The +1 is just a kludge to adjust your range to 1 to 11

Mark Ursino · Accepted Answer · 2012-02-28 16:17:40Z

1

You can use Collection class for sort the ArrayList:

ArrayList<Integer> al =new ArrayList<Integer>();

Collections.sort(al , new Comparator<Integer>()
{
  public int compare(Integer a, Integer b) {
    return a.compareTo(b) ;
  }
});

So now the al has a sorted list.

edited Feb 28, 2012 at 16:17

Mark Ursino

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answered Feb 28, 2012 at 12:44

neeraj

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Comments

David Kroukamp · Accepted Answer · 2012-07-23 15:52:45Z

0

It will do:

Collections.sort();

edited Jul 23, 2012 at 15:52

David Kroukamp

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answered Feb 28, 2012 at 12:40

Kunal Vashist

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