Randomly choose int number

Question

I do have the following code for taking random int numbers

for (int i=1;i<=5;i++) {
  int rand= new Random().nextInt(10);
  Log.d("Ramdom number", String.valueOf(rand));
}

The problem is that I dont want random numbers to be repeated, mean when I run this code it gives to me 5 numbers but two of them at least repeats. Any advice?

Answer 1

For a small range of numbers to choose from, this should do the trick:

ArrayList<Integer> numbers = new ArrayList<Integer>();
for (int i = 0; i < 20; ++i) {
    numbers.add(i);
}
Collections.shuffle(numbers);
for (int i = 0; i < 5; ++i) {
    Log.d("Random number", numbers.get(i).toString());
}

Answer 2

The problem is that you are creating a Random object within the loop. If the loop is a 'tight', as in this case, the Random object will be seeded with the same value. Moving the Random object initialization outside the loop should do the trick.

Random r = new Random();
for (int i=1;i<=5;i++) {
            int rand= r.nextInt(10)

            Log.d("Ramdom number", String.valueOf(rand));  

    }

EDIT:

This should work (at least it did for me)

public static Integer[] getRangedInt(int maxRange, int numCount)
{
    if (maxRange < numCount)
    {
        throw new Exception("maxRange cannot be smaller than numCount");
    }

    Set<Integer set = new HashSet<Integer>();
    Random r = new Random();
    while (Set.size() < numCount)
    {
        int random = r.nextInt(maxRange);  
        while (!set.add(random))
        {
            random = r.nextInt(maxRange); 
        }
    }

    return set.toArray(new Integer[set.size()]);
}

Answer 3

final int maxnumbers = 5;
final int maxvalue = 10;
final Random generator = new Random();
Set<Integer> numbers = new HashSet<Integer>();
while(numbers.size() < maxnumbers){
    numbers.add(random.nextInt(maxvalue));
}

After this loop you should have maxnumber non-repeating random numbers between 0 and maxvalue in the set numbers. You have to watch out so you don't get too many iterations when using this method, i.e. generating 9999 non-repeating numbers out of 10000 would probably take a long time.

Another more scalable version would be to have a list of numbers:

List<Integer> numbers = new ArrayList<Integer>();
for(int i = 0; i<maxvalue; i++){ numbers.add(i); }
Collections.shuffle(numbers);
List<Integer> randomnums = numbers.subList(0, maxnumbers);

Answer 4

So what you're looking for isn't a list of random numbers, it's a list of 30 numbers randomly ordered.

One way is to generate a list of all possible values, then order them randomly, then peel them from the front of the list as needed. Here's some pseudocode:

for(int i=1; i<=30; i++) {
    double r = rand();
    while (null != aTreeSet.get(r)) r = rand();
    aTreeSet.put(r, i);
}

where rand() returns some random value (not the 1-30 you seek, that's i) perhaps between 0 and 1 and aTreeSet is what you think.

The loop prevents sadness in the unlikely event of a dup being returned by rand().

To use this, pull values from aTreeSet in sorted order.

edit - Gross solution

Another way is to generate the 1-30 value, and if it isn't already in a "I have seen this" Set, add it and return the value. if it is there, generate a new random number. Repeat until an unused number is discovered. This performs poorly, relatively speaking, for the last few values. For 30 values on modern processors, it will get done in milliseconds, of course. If your max value was 1,000 instead of 30, I'd start getting concerned.

Answer 5

I think you need a SET of random numbers. This hint should suffice.

If not, then please comment.

Answer 6

You can maintain one list of generated numbers

boolean flag=false;
Vector<int> vec = new Vector<int>();
for (int i=1;i<=5;i++) {
        flag=false;
        int rand= r.nextInt(10);
        for(int j=0;j<vec.size();j++)
        {
             if(vec.get(j)==rand)
             { 
                flag=true;
                break;
             } 
        }
        if(flag)
        {
            continue;
        } 
        else
        {
             Log.d("Ramdom number", String.valueOf(rand));
             vec.add(rand);

        }
}

You can maintain a vector of generated numbers and check

Is that number already generated then generate new one

else display this number

Answer 7

What you want is a random combination, use a Hash table to avoid repetitions

From the top of my head the code should be something like:

    Ramdom r = new Random();
    Hashtable<Integer, Integer> h = new Hashtable<Integer, Integer>();
    while( h.keys().size() < 5 ) {
            int i = r.nextInt(10);
            h.put(i,i);
    }
    Integer[] k = (Integer[]) h.keySet().toArray();

The line

    h.put(i,i);

just overrides the value if it is repeated, so only different drawn numbers will have entries in the hash table.

Answer 8

You can save your generated numbers in a Set and use the random number only if it is not in the set

Random r = new Random();
    Set<Integer> generatedNumbers = new HashSet<Integer>();
    for(int i = 1;i<=5;i++)  {
      int rand = r.nextInt(10)
      if (!generatedNumbers.contains(rand)) {
        Log.d("Ramdom number", String.valueOf(rand));
        generatedNumbers.add(rand);
      } 
    }