How to exclude number in random number in Java

Question

So I will get a random number between 1 to 9999, but I want to exclude 1111, 3333, 4444, 7777 and is that use while loop?

Random r = new Random();

int x = r.nextInt(9999);

while (x == 1111 || x==3333){
    x = r.nextInt(9999) + 1;
}

You don't declare x before the while... This code won't work. You'll need to check `random` — OneCricketeer, Nov 26 '15 at 17:46
It is unclear what are you asking! Please edit your question to include exactly what do you want? — Am_I_Helpful, Nov 26 '15 at 18:34
@kappa If the solution below solves your problem, you can accept the solution by clicking the hollow tick beside the solution. You get 2 rep points in return. — user3437460, Nov 26 '15 at 19:05

score 2 · Answer 1 · answered Nov 26 '15 at 18:55

2

Your solution is fine. You may also use a do-while loop. So that it will generate a random number first, then check.

Random rnd = new Random();
int x=0;
do{
    x = rnd.nextInt(9999)+1;
}while(x==1111 || x==3333 || x==4444 || x==7777);

answered Nov 26 '15 at 18:55

user3437460

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@kappa `but while (x == 1111 || x == 3333 || x == 4444 || x == 7777) { x = r.nextInt(9999); } `The code will work. You can test it from 1-9 and exclude 1,3,4,7. – user3437460 Nov 26 '15 at 19:03

Jorge Campos · Answer 2 · 2015-11-26T18:59:08.500

0

Yes, in this case the while means that it will continue to process a new random number until it is different from 1111 or 3333 on your code.

There is nothing wrong with your code, only minor things.

1 - Fix your while to attend your condition:

while (x == 1111 || x == 3333 || x == 4444 || x == 7777)

2 - The method nextInt process a random number from 0 (inclusive) to the given number exclusive as the Javadocs states (thanks to @user3437460):

Returns a pseudorandom, uniformly distributed {@code int} value between 0 (inclusive) and the specified value (exclusive), drawn from this random number generator's sequence.

So:

Random r = new Random();
int x = r.nextInt(9999)+1;
while (x == 1111 || x == 3333 || x == 4444 || x == 7777) {
    x = r.nextInt(9999) + 1;
}
System.out.println(x);

edited Nov 26 '15 at 18:59

answered Nov 26 '15 at 18:45

Jorge Campos

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r.nextInt(9999) will generate 0 - 9998. Are you sure your solution is correct? – user3437460 Nov 26 '15 at 18:50
but while (x == 1111 || x == 3333 || x == 4444 || x == 7777) { x = r.nextInt(9999); } still have chance to get 1111 – kappa Nov 26 '15 at 18:51
you can use a do { ... } while(....); which will remove the duplicate r.nextInt(9999) + 1 – Thierry Nov 26 '15 at 18:54
yes @Thierry ! You should add your suggestion as an answer! I would gladly upvote it, since it will be a better suggestion than my one :) – Jorge Campos Nov 26 '15 at 18:55
@JorgeCampos Your solution is still incorrect. r.nextInt(10000) will generate 0 - 9999 ! – user3437460 Nov 26 '15 at 18:57
1

@user3437460 Yeah, realize that now! 0 != 1 thanks again. – Jorge Campos Nov 26 '15 at 18:58

How to exclude number in random number in Java

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