Unity C# Random Unique Number

Question

I want to generate random unique number. These are codes:

public int NewNumber(int r) {
    int[] numbers = new int[r];   
    int a = Random.Range (0, r);
    if(numbers.Contains(a)) {
        a = Random.Range (0, r);
    }
    return a;
}

When the NewNumber(5) works ten times, I get result as like: 5,4,3,4,1,2,4,2,5,3

But what I want: 5,3,4,2,1

Where is wrong?

Answer 1

You are resetting your holder every time you call NewNumber method, you need to put your holder outside your method.

int[] numbers = new int[r];   
 public int NewNumber(int r) {
    int a = Random.Range (0, r);
    if(!numbers.Contains(a)) {
        a = Random.Range (0, r);
    }
    return a;
}

It will be better if you use a list instead of array;

List<int> numbers = new List<int>();
public int NewNumber(int r) {
   int a = Random.Range (0, r);
   if(!numbers.Contains(a)) {
       numbers.Add(a);
   }
   return a;
}

Answer 2

If there's only one number within the range that you don't want to be returned, the solution needs no array. Instead simply repeat creating a new random number if the random number happens to be the one you don't want.

public int RandomNumberInRangeExcluding(int range, int excludedNumber)
{
    int r = excludedNumber;
    while (r == excludedNumber)
    {
        r = Random.Range(0, range);
    }
    return r;
}

Answer 3

nexx is right but you're still going to get duplicates unless you wrap this in a loop, something like...

List<int> numbers = new List<int>();

public int NewNumber(int r) {

    int a = 0;

    while(a==0){    
        a = Random.Range (0, r);
        if(!numbers.Contains(a)) {
            numbers.Add(a);
        }else{
            a=0;
        }
    }
    return a;
}

Answer 4

I think it's because you only random once when numbers contains a . Try changing if to while so the random continues until numbers doesn't contain a

int[] numbers = new int[r];   
public int NewNumber(int r) {
    int a = Random.Range (0, r);
    while(numbers.Contains(a)) { // change this to while
        a = Random.Range (0, r);
    }
    return a;
}

Answer 5

this achieves the same result, without waiting for Random to hopefully spit out whichever numbers are needed

    static void Main(string[] args)
    {
        List<int> rands = GetRandoms(10);
    }

    static IEnumerable<int> GetRandoms(int size)
    {
        List<int> rands = new List<int>();

        for (int i = 1; i <= size; i++)
        {
            rands.Add(i);
        }

        return rands.OrderBy(i => Guid.NewGuid());
    }