Creating a 4 digit Random Number using java with no repetition in digits

Question

I wrote a code using java to create a random 4 digit number with no repetition of digits, the code I wrote is given below :-

Random r = new Random();
d1 = r.nextInt(9);
d2 = r.nextInt(9);
d3 = r.nextInt(9);
d4 = r.nextInt(9);
while(d1==d2||d1==d3||d1==d4||d2==d3||d2==d4||d3==d4)
{
    if(d1==d2||d2==d3||d2==d4)
    {
        d2 = r.nextInt(9);
    }
    if(d1==d3||d2==d3||d3==d4)
    {
        d3 = r.nextInt(9);
    }
    if(d1==d4||d2==d4||d3==d4)
    {
        d4 = r.nextInt(9);
    }
}   
System.out.println(d1+""+d2+""+d3+""+d4);

here are the test cases(generated from System.out.println(R1+""+R2+""+R3+""+R4);) are as following :-

 0123 |  OK as required
 1234 |  OK as required
 2123 |  not OK because 2 is present more than one time 
 9870 |  OK as required
 0444 |  not OK because 4 is present more than one time

Now My question here is, that if there is some better way to do this. If I could enhance it in some way?

Answer 1

A couple of approaches:

1. Use a Set to hold the digits and keep adding random digits until the set has four values in it.

2. Create an array with values 0-9 in it. Shuffle the array and take the first four values.

If performance matters, you will want to try a couple of different methods and see which is faster.

Answer 2

Create a list of integers from 0 to 9, shuffle it and extract the first 4.

public static void main(String[] args) {
    List<Integer> numbers = new ArrayList<>();
    for(int i = 0; i < 10; i++){
        numbers.add(i);
    }

    Collections.shuffle(numbers);

    String result = "";
    for(int i = 0; i < 4; i++){
        result += numbers.get(i).toString();
    }
    System.out.println(result);
}

There's some ugly string-to-int conversing going on, but you get the idea. Depending on your use case you can see what is needed.

Answer 3

Here's my approach, even though it uses a lot of string parsing but no data structure:

 static int generateNumber(int length){
            String result = "";
            int random;
            while(true){
                random  = (int) ((Math.random() * (10 )));
                if(result.length() == 0 && random == 0){//when parsed this insures that the number doesn't start with 0
                    random+=1;
                    result+=random;
                }
                else if(!result.contains(Integer.toString(random))){//if my result doesn't contain the new generated digit then I add it to the result
                    result+=Integer.toString(random);
                }
                if(result.length()>=length){//when i reach the number of digits desired i break out of the loop and return the final result
                    break;
                }
            }

            return Integer.parseInt(result);
        }

Answer 4

Use a Set maybe?

Random r = new Random();
Set<Integer> s = new HashSet<Integer>();
while (s.size() < 4) {
    s.add(r.nextInt(9));
}