Generate 6 different numbers from 1 to 10 in Java

Question

I wrote down my code at first and then it cannot ensure that it is 6 different number.

public class JavaTest{
public static void main(String[] args){
    int[] list = new int[6];
    int number = (int)(Math.random() * 10 + 1);

    for (int i = 0; i < list.length; i++){
        list[i] = number;
        number = (int)(Math.random() * 10 + 1);
        for (int j = 1; j < i; j++){
            number = (int)(Math.random() * 10 + 1);
            if (number == list[j])
                number = (int)(Math.random() * 10 + 1);
        }
    }

    for (int i = 0; i < list.length; i++){
        System.out.print(list[i] + " ");
    }
}
}

I think something wrong with my inner for loop j, but I don't know how to improve it. Can anyone help me modify it? Thanks a lot!!

Bubletan · Answer 1 · 2015-03-22T18:24:15.003

9

You could possibly do something like:

List<Integer> list = Arrays.asList(1, 2, 3, 4, 5, 6, 7, 8, 9, 10);
Collections.shuffle(list);

And then take the 6 first numbers.

Without using collections:

int[] list = new int[6];
boolean[] used = new boolean[10];
for (int i = 0; i < 6; i++) {
    int number = (int) ((10 - i) * Math.random());
    for (int a = 0; a <= number; a++) {
        if (used[a]) {
            number++;
        }
    }
    list[i] = number + 1;
    used[number] = true;
}

edited Mar 22 '15 at 18:24

answered Mar 22 '15 at 17:38

Bubletan

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Actually I know how to solve it in this way, but I just want to try another way to solve it. Thanks a lot! :) – yolam Mar 22 '15 at 17:51
@yolam Added a version without collections. Edit: Actually it's not completely random. I'll edit it. – Bubletan Mar 22 '15 at 17:53
Fixed, should be fine now. – Bubletan Mar 22 '15 at 18:24
Thank you very much!! :) – yolam Mar 22 '15 at 18:27

score 3 · Answer 2 · answered Mar 22 '15 at 17:42

3

You can just use Set. Set will ensure that there will be 6 different numbers.

public static void main(String[] args) {
    Random random = new Random();
    Set<Integer> numbers = new HashSet<>(6);

    while (numbers.size() < 6) {
        numbers.add(1 + random.nextInt(10));
    }

    System.out.println(numbers);
}

answered Mar 22 '15 at 17:42

Aleksander Mielczarek

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Set does not keep an order of the added elements. He could as well just go ahead and add numbers from 1 to 6 to produce the effect like you provided. – Sebastian Brudzinski Mar 22 '15 at 17:52
Task was to generate 6 different numbers, any word about their order. – Aleksander Mielczarek Mar 22 '15 at 17:56
I am just a beginner and cannot understand Set now, sorry. I will find materials related to Set later and thanks a lot! :) – yolam Mar 22 '15 at 17:57

score 0 · Answer 3 · edited May 23 '17 at 12:08

0

You can do something like that:

public static int randInt(int min, int max) {

    Random rand = new Random();

    int randomNum = rand.nextInt((max - min) + 1) + min;

   return randomNum;
}

Source: How do I generate random integers within a specific range in Java?

edited May 23 '17 at 12:08

Community

1
1

answered Mar 22 '15 at 17:41

NDY

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Generate 6 different numbers from 1 to 10 in Java

3 Answers3