For random number

Question

I have generate a random number from 1-100 but not the duplicate number will appear(eg.if 10 comes ones then it will not come again.) (this is not working)

 final Random random = new Random();
    ArrayList<Integer> arrayList = new ArrayList<Integer>();

    while (arrayList.size() < 6) { // how many numbers u need - it will 6
        int a = random.nextInt(91)+1; // this will give numbers between 1 and 90.

        if (!arrayList.contains(a)) {
            arrayList.add(a);
        }
    }

and how can I used that generated appear at the bottom of the screen so that what number is done user must know If 1-100 all the number is done then no number is generated

Can anyone help me with this

check this link : [https://stackoverflow.com/a/8115744/7271027](https://stackoverflow.com/a/8115744/7271027) — Dinesh, Apr 15 '20 at 08:41
"this is not working" in what way doesn't it work? [Looks to work to me](https://ideone.com/ABjnUU). — Andy Turner, Apr 15 '20 at 08:45

score 1 · Accepted Answer · answered Apr 15 '20 at 08:42

1

Try this may be help you

Random rng = new Random(); // Ideally just create one instance globally
List<Integer> generated = new ArrayList<Integer>();
for (int i = 0; i < numbersNeeded; i++)
{
    while(true)
    {
        Integer next = rng.nextInt(max) + 1;
        if (!generated.contains(next))
        {
            // Done for this iteration
            generated.add(next);
            break;
        }
    }
}

answered Apr 15 '20 at 08:42

Jignesh Mayani

6,937
1
20
36

yet working but not optimal! – AouledIssa Apr 15 '20 at 08:45
i am using this and this will helpful for me – Jignesh Mayani Apr 15 '20 at 08:51

score 0 · Answer 2 · answered Apr 15 '20 at 08:41

0

Here try this.

Integer[] arr = new Integer[100];
    for (int i = 0; i < arr.length; i++) {
        arr[i] = i;
    }
    Collections.shuffle(Arrays.asList(arr));

answered Apr 15 '20 at 08:41

Parth

791
7
17

duplicate of https://stackoverflow.com/a/21213707/2057782 – AouledIssa Apr 15 '20 at 08:46
well i didn't search for the answer. anyways thank you for your suggestion :) – Parth Apr 15 '20 at 08:48
No worries! but that should be a common practice. – AouledIssa Apr 15 '20 at 08:56

For random number

2 Answers2