Creating unique random numbers in Java

Question

I was making a Random number (sort of a guessing game) and have come up with the ff. code to generate 10 one or two-digit numbers(1 or 10 up to 40):

public void generate()
{
    for(int i=0; i<=1; i++)
    {
        for(int l=0; l<10; l++)
        {
            Random rdm=new Random();
            arr[l] = rdm.nextInt(range)+1;

        }
    }
}

However, this code only answers the need to generate 10 random one or two-digit numbers. I need to make this program generate unique random numbers. How can I do that?

sorry for the late update... what I want to do with this program is that if the array contains a duplicate, that duplicate would be replaced with a unique one...

==============SOLVED================

NEW PROBLEM:

HashSet set=new HashSet();
Random random=new Random();

while(set.Size()<10)
{
    set.add(random.nextInt(range)+1);
}

lbtest.setText(set.toString());
bgen.setEnabled(false);
gametext.setText("");

As requested by ggrigery:

here's the updated code in reference to ggrigery's suggestion.

you can use a structure, ArrayList for example, to store previously generated numbers and then check to see if the new number is in that list. — twain249
– twain249, Commented Mar 13, 2012 at 14:07
See also this question, which has an almost-working implementation. — Ry-
– Ry- ♦, Commented Mar 13, 2012 at 14:14
+1 to get this back to 0, not sure why it was voted down as it seems like a valid question to me. You want to use a Set, see my answer below :) — ggrigery
– ggrigery, Commented Mar 13, 2012 at 16:02
@Paul I think it's being used to mean following, which would be unusual but I suppose valid. — AakashM
– AakashM, Commented Mar 13, 2012 at 16:25

Peter Lawrey · Accepted Answer · 2012-03-13 14:27:58Z

5

Another option is to use shuffle.

List<Integer> all = new ArrayList<>();
for(int i=1;i<=range;i++) all.add(i);
Collections.shuffle(all);
List<Integer> selected = all.subList(0, 10);

If you are selecting every element, it can take a long time to find the last random value if you are discarding duplicates. This approach takes the same amount of time whether you select one or all elements.

answered Mar 13, 2012 at 14:27

Peter Lawrey

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Comments

ggrigery · Accepted Answer · 2012-03-13 16:18:18Z

1

    HashSet<Integer> set = new HashSet<Integer>();
    Random random = new Random();
    int i = 0;

    while(set.size() < 10){
        set.add(random.nextInt(40) + 1);
        i++;
    }

    System.out.println(set);
    System.out.println(i);

edited Mar 13, 2012 at 16:18

answered Mar 13, 2012 at 15:58

ggrigery

4113 silver badges14 bronze badges

9 Comments

cryzone Over a year ago

QUESTION: how do i display the elements from the set(HashSet)? does it replace the duplicates once found within the set?

ggrigery Over a year ago

Sets cannot contain duplicates by nature. Set.add() returns a boolean; true if the element does not exist and it was added to the set, false if the addition "failed". If you add a counter to that code, you will see that the loop generally executes more than 10 times.

ggrigery Over a year ago

To answer your display question, try System.out.println(set). The default toString() method will print out the elements like [element1, element2, ....]. I have updated the code with a counter, try running it and see what happens.

ggrigery Over a year ago

try setText(set.toString);. In my code above, the toString() is implicitly called. If you want to pass a String, you will need to explicitly call the toString() method.

ggrigery Over a year ago

Sorry, I didn't completely read the error. The error does not appear to be related to setText(). Can you update the code you are using now in your original question and we can look if you have any syntax issues?

|

viktor · Accepted Answer · 2012-03-13 14:19:10Z

0

Set guarantees unique elements. Then you may take result and convert it in whatever collection or array you want.

public List<Integer> generate() {
   Random random = new Random();
   Set<Integer> set = new LinkedHashSet<Integer>();
   for(int i = 0; i < numberCount; i++) {
      set.add(random.nextInt(range));
   }
   return new ArrayList<Integer>(set);
}

answered Mar 13, 2012 at 14:19

viktor

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3 Comments

Ry- Over a year ago

I don't really know Java's structures, but I think that won't keep generating numbers until there are numberCount; it'll just eliminate the duplicates.

cryzone Over a year ago

Q: does this code generate 10 one or two-digit random numbers?

ggrigery Over a year ago

I didn't see this answer before posting my own. Mine is the same concept, but mine will guarantee that the set is 10 numbers in the range of 1-40 (inclusive).

Th0rndike · Accepted Answer · 2012-03-13 14:21:03Z

0

Do a search inside the array to know if the value is there:

private boolean isInsideArray(int randomNumber, int[] arr){
    for(int x = 0; x < arr.length; x++){
        if(arr[x] == randomNumber) return true;

    }
    return false;
}

Just call this function when you want to know if a number is inside the array.

NOTE: this code is untested, it is just an example to help you out. Also not the most elegant solution, there are many methods in Java that can help you accomplish this without cycling the entire array.

edited Mar 13, 2012 at 14:21

answered Mar 13, 2012 at 14:14

Th0rndike

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2 Comments

Ry- Over a year ago

I think you meant to put the return false after the loop. Also, if you use an ArrayList, you get ArrayList.contains, and an ArrayList is much better for this situation.

Th0rndike Over a year ago

the guy is just doing homework to learn the loops. Agreed that an arraylist is better, this solution is better for what he is learning today. Don't worry, he will learn the ArrayLists very soon :)

arkk · Accepted Answer · 2012-03-13 14:26:27Z

0

public void generate() {
    Random rdm = new Random();
    int i = 0;
    while (i < 10) {
        int j = rdm.nextInt(range) + 1;
        if (!arrayContains(j)) {
            arr[i] = j;
            i++;
        }
    }   
}

public boolean arrayContains(int i) {
    for (int k = 0; k < arr.length; k++) {
        if (arr[k] == i)
            return true;
    }
    return false;
}

This should do the job. Could be more elegant though.

answered Mar 13, 2012 at 14:26

arkk

1

2 Comments

cryzone Over a year ago

@arkk_what does this mean if(!arrayContains(j))?

Edwin Buck Over a year ago

@cryzone, if(!arrayContains(j)) can be read in English as "if arrayContains(j) is not true".

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