How to generate array of unique values

Question

I'm trying to generate an array of unique values but my first number is always 0, how can I fix this?

Random r = new Random();
Scanner in = new Scanner(System.in);
int generatedNumbers = 0;
int winningSum = 0;
int[] list = new int[10];

for (int number = 1; number < list.length; ++number) {
    list[number] = r.nextInt(99);

    for (int uniqueNumber = 0; uniqueNumber < number; ++uniqueNumber) {
        if (list[uniqueNumber] == list[number]) {
            list[number] = r.nextInt(99);
            uniqueNumber = 0;
        }
    }
}
int random_first = r.nextInt(99);)
for (int element : list) {
    System.out.println(element);
}

Please read: How to debug small programs

Turing85
– Turing85

2021-11-12 19:59:52 +00:00
Commented Nov 12, 2021 at 19:59 — Turing85
– Turing85, Commented Nov 12, 2021 at 19:59

Ben C · Accepted Answer · 2021-11-12 21:05:22Z

1

you need to start from zero, not one, in your for loop

Random r = new Random();
int[] list = new int[50];
for(int index = 0; index < list.length; index++) {
    int newNum = r.nextInt(99);
    outer:while(true){
        for(int i = 0;i<index;i++){
            if(list[i]==newNum){
                newNum = r.nextInt(99);
                continue outer;
            }
        }
        break;
    }
    list[index] = newNum;
    
}
for (int element: list) {
    System.out.println(element);
}

edited Nov 12, 2021 at 21:05

answered Nov 12, 2021 at 20:25

Ben C

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4 Comments

user17223316 Over a year ago

It sometimes has non-unique results.

Ben C Over a year ago

I fixed it but be careful because it can get stuck if it cannot find another number available so obviously make sure the list is small-ish

user17223316 Over a year ago

I think i <list.length may be i <index as OP does.

Ben C Over a year ago

that would be more efficient yes

djm.im · Accepted Answer · 2021-11-12 21:34:08Z

0

In your approach, you have three mistakes (that I noticed).

The first, this loop for (int number = 1; number < list.length; ++number) should start with zero. So, for (int number = 0; number < list.length; ++number).

The second, inner check for (int uniqueNumber = 0; uniqueNumber < number; ++uniqueNumber) { fails here uniqueNumber = 0; It should be uniqueNumber = -1.

And finally, you don't need this line int random_first = r.nextInt(99);) after outer for loop.

Bonus. For easier debugging, you can use a trick. Replace 99 with 10. Why? When you generate a list all values in it should be unique.

Random r = new Random();
int[] list = new int[10];

int bound = 10;
for (int number = 0; number < list.length; ++number) {
    list[number] = r.nextInt(bound);

    for (int uniqueNumber = 0; uniqueNumber < number; ++uniqueNumber) {
        if (list[uniqueNumber] == list[number]) {
            list[number] = r.nextInt(bound);
            uniqueNumber = -1;
        }
    }
}

edited Nov 12, 2021 at 21:34

answered Nov 12, 2021 at 21:14

djm.im

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1 Comment

user17223316 Over a year ago

The body of the if statement may be --number; break;.

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