generate random numbers with no repeat in c#

Question

How can I generate random numbers with no repeat in C#. I have one array and I want to fill every room with random numbers from 0 to 9. Each room shoud have diffrent numbers

Answer 1

100 % right

   public int[] UniqeRandomArray(int size , int Min , int Max ) {

        int [] UniqueArray = new int[size];
        Random rnd = new Random();
        int Random;

        for (int i = 0 ; i < size ; i++) {

            Random = rnd.Next(Min, Max);

            for (int j = i; j >= 0 ; j--) {

                if (UniqueArray[j] == Random)
                { Random = rnd.Next(Min, Max); j = i; }

            }

            UniqueArray[i] = Random;

        }

        return UniqueArray;

    }

// Notice to be unique [Max - Min > size] NOT equal

Answer 2

This will create a unique range of 1 to rangeEx inclusive. The next two lines create a random number Generator and orders the IEnumerable range with a randome number. this is then called with ToArray and returned!

   private int[] RandomNumber(int rangeEx)
    {
        var orderedList = Enumerable.Range(1, range);
        var rng = new Random();
        return orderedList.OrderBy(c => rng.Next()).ToArray();
    }

Answer 3

Your problem is that you are creating the Random object in every loop. The Random object must be created only once. Try this instead:

Random rnd = new Random(); // <-- This line goes out of the loop        
for (int i = 0; i < 20; i++) {
    int temp = 0;
    temp = rnd.Next(0, 9);
    page[i] = temp;
}

Answer 4

        ArrayList page=new ArrayList();
        int random_index;            
        random rnd = new Random();

        for (int i = 0; i < 20; i++)
        {
            do                            
            {
                random_index = rnd.Next(10);
                if (!(page.Contains(random_index)))
                                break;
            } while (page.Contains(random_index));
            page.Add(random_index);
        }

Answer 5

    public Form1()
    {
        InitializeComponent();
    }
    int A, B;
    string Output;
    private void Form1_Load(object sender, EventArgs e)
    {
        for (int i = 0; i < 20; i++)
        {
            while (A == B)
            {
                Random r = new Random();
                A = r.Next(1, 6);
            }
            Output = Output + A;
            B = A;
        }
        textBox1.Text = Output;
    }

Output: 24354132435213245415 (not repeating)

Answer 6

With such a small list of numbers to choose from you can simply generate a list that contains all of them and then shuffle them.