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I want to generate a random number of any particular digit (say 5, i.e., 10000 to 99999) in java. How can I do that?

sampathsris
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sbanerjee
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5 Answers5

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Create a random integer in range 0,10^k - 10^(k-1), and add 10^(k-1) - in your example. range 0,90000, and add 10000 to it.

Use the Random.nextInt(int) method for it.

Something like:

new Random().nextInt(90000) + 10000
amit
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  • This does not support negative ranges. Instead use `rand.nextInt(upper - lower) + lower;` – Upio Sep 04 '14 at 08:51
  • @Upio You are answering a generic question of "how to produce integer in a specific range". This answer aims to show how to create a k digits integer, which is what the OP has asked for. – amit Sep 04 '14 at 08:52
  • You're right. I misinterpreted what he meant by any digit. :) – Upio Sep 04 '14 at 08:53
  • but in case of a 9 or higher digit number what should i do? Integer/int variable will overflow right? please help to make it generic. – sbanerjee Sep 04 '14 at 11:01
  • @sbanerjee You can also use [`nextLong() % number`](http://docs.oracle.com/javase/7/docs/api/java/util/Random.html#nextLong()) to get a long in range (0,number). If your numbers are bigger than longs, you are going to need a [BigInteger](http://docs.oracle.com/javase/7/docs/api/java/math/BigInteger.html). You can create several longs until you fill up k-1 digits (including leading zeros), and then geenrate a number in range [1,9], and make it the most significant digit. – amit Sep 04 '14 at 11:49
0

Simple.

  1. generate a random number upto 89999
  2. Add 10000 to each generated number.
TheLostMind
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public int generateRandom() {
    Random r = new Random( System.currentTimeMillis() );
    return 10000 + r.nextInt(20000);
}

this is from here

or :

for (int i = 0; i < 10; i++) {  
     System.out.println(Math.round(Math.random() * 89999) + 10000);  
}
Community
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user3145373 ツ
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0

The following should work: 

public int getRandomNumber(int digits)
{
    assert digits > 0; //added because of comment
    BigInteger tenPower = new BigInteger("10");
    return new Random().nextInt((tenPower.pow(digits).subtract(tenPower.pow(digits-1))).intValue()) + tenPower.pow(digits-1).intValue();
}

Output:

Digits: 1
  - Random nextInt Param: 9
  - Adding: 1
  Random number: 1
Digits: 2
  - Random nextInt Param: 90
  - Adding: 10
  Random number: 66
Digits: 3
  - Random nextInt Param: 900
  - Adding: 100
  Random number: 377
Digits: 4
  - Random nextInt Param: 9000
  - Adding: 1000
  Random number: 1734
Digits: 5
  - Random nextInt Param: 90000
  - Adding: 10000
  Random number: 96602

So it creates the numbers like this:

1...9
10...99
100...999
1000...9999
10000...99999
EpicPandaForce
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  • So it's worth noting that for 1 digit, it doesn't generate a `0`. Other than that, it's cool. You can also change it to `long` or keep it as `BigInteger` if you want. – EpicPandaForce Sep 04 '14 at 08:51
  • Are you serious? What if i pass 10 or -5 as parameter? – user1516873 Sep 04 '14 at 08:52
  • @user1516873 I'm not sure I can understand what `negative digit count` even means, or even a number with `0 digits`. I could just add an assert to prevent that from happening, if you're so inclined. As I said, if you want, you can keep it as a `BigInteger` though if you want to handle large numbers. – EpicPandaForce Sep 04 '14 at 08:55
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Here is how you can handle negative and positive ranges. Compute a random number within the size of the range and then add the lower bound. E.g. if the range is 100-150, then compute a number between 0 and 50 and add 100. This works for both positive and negative ranges.

import java.util.Random;

public class ArbitraryRange {

     public static void main(String []args){
        int start = -10000;
        int end = -5000;

        Random rand = new Random();
        int r = rand.nextInt(end - start) + start;

        System.out.println(r);
     }
}
Upio
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