How to generate a random number that is unique each time

Question

Hey I am trying to create a program that generates a random number that is only allowed to be used once. Sorry if that's confusing I'll try to explain. I want to have a program generate numbers from 1-100, but say if it generates 33, then for the rest of the number generation process 33 cannot be generated again, So I should end with exactly 100 different numbers. Any help is really appreciated, Thank you.

Here's my attempt so far

public class Seed {

    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int a =0; 
        for (int i =0; i <20;i++) {
             a = (int) Math.ceil(10 * Math.random())  ;
            System.out.println(a);
            int x = a;
            System.out.println("This is x: " + x);

            if (x == a )
            {
                a = (int) Math.ceil(10 * Math.random())  ;
            }
        }
    }
}

Use a `Set`, add each new number to it. Keep generating a number until it no longer exists within the `Set` — MadProgrammer, Oct 22 '14 at 02:57
@MadProgrammer that would work, although it's gonna get really slow when you are looking for the last remaining numbers — Greg Case, Oct 22 '14 at 03:04
possible duplicate of [Creating random numbers with no duplicates](http://stackoverflow.com/questions/4040001/creating-random-numbers-with-no-duplicates) — Hot Licks, Oct 22 '14 at 03:04

score 11 · Accepted Answer · answered Oct 22 '14 at 02:59

Precompute a List containing all the required numbers, then shuffle it and remove elements each time you need to generate a new number. Eg:

List<Integer> numbers = new ArrayList<Integer>(100);

for (int i = 0; i < 100; ++i)
  numbers.add(i);

Collections.shuffle(numbers);

int pick = numbers.remove(0);
int pick2 = numbers.remove(0);

How to generate a random number that is unique each time

1 Answers1