我如何才能使随机数不再在数组中重复？

Question

listBox1.Items.Clear();
int[] sayısal = new int[6];
Random rastgele = new Random();
for (int i = 0; i < 6; i++)
{
   do
   {
      sayısal = rastgele.Next(1, 50);
   }
   while (listBox1.Items.IndexOf(sayısal) != -1);
      listBox1.Items.Add(sayısal);
}

When I did like this, I take an error that calls

"Cannot implicitly convert type 'int' to 'int[]' "

in line "sayısal = rastgele.Next(1, 50);" . What can I do for it?

Answer 1

You can generate sequence 1..50 and shuffle it (ie sort by random value):

Random rastgele = new Random();
int[] sayısal = Enumerable.Range(1, 50) // generate sequence
                          .OrderBy(i => rastgele.Next()) // shuffle
                          .Take(6) // if you need only 6 numbers
                          .ToArray(); // convert to array

---

Your code is not working, because you are trying to assign generated item to array variable.

sayısal = rastgele.Next(1, 50);

It should be instead:

do {
   sayısal[i] = rastgele.Next(1, 50);
} while(listBox1.Items.IndexOf(sayısal[i]) != -1);

---

As I already pointed in comments, it's better to separate UI logic and array generation. Ie

// generate array (optionally move to separate method)
int itemsCount = 6;
int[] items = new int[itemsCount]; // consider to use List<int>
Random random = new Random();
int item;

for(int i = 0; i < itemsCount; i++)
{
   do {
      item = random.Next(1, 50);
   } while(Array.IndexOf(items, item) >= 0);

   items[i] = item;
}

// display generated items
listBox1.Items.Clear();
for(int i = 0; i < items.Length; i++) // or use foreach
    listBox1.Items.Add(items[i]);

Answer 2

Because Random.Next method returns an int , not int[] . And there is no implicit conersation from int[] to int .

Return Value
Type: System.Int32
A 32-bit signed integer greater than or equal to minValue and less than maxValue; that is, the range of return values includes minValue but not maxValue. If minValue equals maxValue, minValue is returned.

If you want to fill your array, you can use Enumerable.Range like lazyberezovsky mentioned .

Answer 3

This method takes an integer array and randomly sorts them. So fill an array with a loop then use this to randomly sort the array. You should credit one of the others as they were first to post with valid answers. I just thought another way to do this would be good.

amount is the amount of times you want the array to randomize. The higher the number the higher the chance of numbers being random.

    private Random random = new Random();

    private int[] randomizeArray(int[] i, int amount)
    {
        int L = i.Length - 1;
        int c = 0;
        int r = random.Next(amount);
        int prev = 0;
        int current = 0;
        int temp;

        while (c < r)
        {
            current = random.Next(0, L);
            if (current != prev)
            {
                temp = i[prev];
                i[prev] = i[current];
                i[current] = temp;
                c++;
            }
        }
        return i;
    }

Answer 4

Be careful in your choice of data structures and algorithms, pick the wrong one and you'll wind up with O(n^2). A reasonable solution IMHO is a typed hash table (ie dictionary) which will give you O(n):

Random rnd = new Random();              
        var numbers = Enumerable
            .Range(1, 1000000)
            .Aggregate(new Dictionary<int, int>(), (a, b) => {
                int val;
                do {val = rnd.Next();} while (a.ContainsKey(val));
                a.Add(val, val);
                return a;
            })
            .Values
            .ToArray();

Still not ideal though as performance depends on the array size being significantly smaller than the set of available numbers and there's no way to detect when this condition isn't met (or worse yet when it's greater, in which case the algorithm will go into an infinite loop).