How do I generate a random integer in C#?

you want to generate unique random numbers in C# so that once a number is picked, it won’t repeat.

Here are a few clean approaches:

Approach 1: Use a HashSet to Track Generated Numbers

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        Random rand = new Random();
        HashSet generated = new HashSet();

        int min = 1, max = 100; // range
        int count = 10;         // how many unique numbers you want

        while (generated.Count < count)
        {
            int num = rand.Next(min, max + 1); // inclusive range
            if (generated.Add(num)) // Add returns false if already exists
            {
                Console.WriteLine(num);
            }
        }
    }
}

• Ensures uniqueness using HashSet.

• Works for any range.

• If you need all numbers in a range, this can get inefficient for large ranges.

Approach 2: Shuffle a List (Best for Full Range)

using System;
using System.Collections.Generic;

class Program
{
    static void Main()
    {
        int min = 1, max = 100;
        List numbers = new List();

        for (int i = min; i <= max; i++)
            numbers.Add(i);

        Random rand = new Random();

        // Fisher-Yates shuffle
        for (int i = numbers.Count - 1; i > 0; i--)
        {
            int j = rand.Next(i + 1);
            (numbers[i], numbers[j]) = (numbers[j], numbers[i]);
        }

        // Now numbers are unique and randomized
        foreach (var num in numbers)
            Console.WriteLine(num);
    }
}

• Generates all numbers in range without duplicates.

• Efficient if you need a complete set of unique randoms.

• Uses more memory for very large ranges.

Approach 3: LINQ One-Liner (Quick & Elegant)

using System;
using System.Linq;

class Program
{
    static void Main()
    {
        Random rand = new Random();
        var uniqueNumbers = Enumerable.Range(1, 100)
                                      .OrderBy(x => rand.Next())
                                      .Take(10);

        foreach (var num in uniqueNumbers)
            Console.WriteLine(num);
    }
}

• Concise.

• Great for small sets.

• Less efficient for huge ranges.

Practice

• If you need all numbers in a range ? use Approach 2 (Shuffle).

• If you need a few unique numbers ? use Approach 1 (HashSet).

you can use Random calss to generate random int number in various way,

  Random rnd1 = new Random();

            var uniqueNumber = new List();
            for (int j = 0; j < 20; j++)
            {
                var num = rnd1.Next();
                if (!uniqueNumber.Contains(num))
                {
                    uniqueNumber.Add(num);
                } 
            }

Note : Next(); has method overloading which will allow you to pass different parameter so you can use according to your requirement.

List listNumbers = new List();
int number;
for (int i = 0; i < 6; i++)
{
  do {
     number = rand.Next(1, 49);
  } while (listNumbers.Contains(number));
  listNumbers.Add(number);
}

You can create a function and add a min-max value that can generate a random number between those min max values

Hello Naimish,

You can use the Random class to create random numbers. Please check below sample code.

Random rnd = new Random();
int month  = rnd.Next(1, 13);  // creates a number between 1 and 12

Thanks

You can use the System.Random class to generate random numbers in C#, and then store the generated numbers in a data structure such as a HashSet to ensure that each number is unique.

int totalNumbers = 10;
HashSet uniqueNumbers = new HashSet();
Random random = new Random();

while (uniqueNumbers.Count < totalNumbers)
{
    int number = random.Next();
    if (!uniqueNumbers.Contains(number))
    {
        uniqueNumbers.Add(number);
    }
}

check this

public class RandomNumberGenerator
{
public Random a = new Random(); 
public List randomList = new List();
int x= 0;
public int GetRandomInt(int min,int max)
{
    x= a.Next(min, max);
    if (!randomList.Contains(x))
        randomList.Add(x);
 return randomList[x];
}

}