Java: Random Long Number in 0 <= X < N Range

Java: random long number in 0 = x n range

Starting from Java 7 (or Android API Level 21 = 5.0+) you could directly use ThreadLocalRandom.current().nextLong(n) (for 0 ≤ x < n) and ThreadLocalRandom.current().nextLong(m, n) (for m ≤ x < n). See @Alex's answer for detail.

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If you are stuck with Java 6 (or Android 4.x) you need to use an external library (e.g. org.apache.commons.math3.random.RandomDataGenerator.getRandomGenerator().nextLong(0, n-1), see @mawaldne's answer), or implement your own nextLong(n).

According to Random documentation, nextInt is implemented as

 public int nextInt(int bound) {
   if (bound <= 0)
     throw new IllegalArgumentException("bound must be positive");

   if ((bound & -bound) == bound)  // i.e., bound is a power of 2
     return (int)((bound * (long)next(31)) >> 31);

   int bits, val;
   do {
       bits = next(31);
       val = bits % bound;
   } while (bits - val + (bound-1) < 0);
   return val;
 }

So we may modify this to perform nextLong:

long nextLong(Random rng, long bound) {
    // error checking and 2^x checking removed for simplicity.
    long bits, val;
    do {
        bits = (rng.nextLong() << 1) >>> 1;
        val = bits % bound;
    } while (bits-val+(bound-1) < 0L);
    return val;
}

Java: Random long value in an interval

If you want range based long values then do the below:

 long LOWER_RANGE = 0; //assign lower range value
 long UPPER_RANGE = 1000000; //assign upper range value
 Random random = new Random();

 long randomValue = LOWER_RANGE + 
                           (long)(random.nextDouble()*(UPPER_RANGE - LOWER_RANGE));

Creating random numbers with no duplicates

The simplest way would be to create a list of the possible numbers (1..20 or whatever) and then shuffle them with Collections.shuffle. Then just take however many elements you want. This is great if your range is equal to the number of elements you need in the end (e.g. for shuffling a deck of cards).

That doesn't work so well if you want (say) 10 random elements in the range 1..10,000 - you'd end up doing a lot of work unnecessarily. At that point, it's probably better to keep a set of values you've generated so far, and just keep generating numbers in a loop until the next one isn't already present:

if (max < numbersNeeded)
{
    throw new IllegalArgumentException("Can't ask for more numbers than are available");
}
Random rng = new Random(); // Ideally just create one instance globally
// Note: use LinkedHashSet to maintain insertion order
Set<Integer> generated = new LinkedHashSet<Integer>();
while (generated.size() < numbersNeeded)
{
    Integer next = rng.nextInt(max) + 1;
    // As we're adding to a set, this will automatically do a containment check
    generated.add(next);
}

Be careful with the set choice though - I've very deliberately used LinkedHashSet as it maintains insertion order, which we care about here.

Yet another option is to always make progress, by reducing the range each time and compensating for existing values. So for example, suppose you wanted 3 values in the range 0..9. On the first iteration you'd generate any number in the range 0..9 - let's say you generate a 4.

On the second iteration you'd then generate a number in the range 0..8. If the generated number is less than 4, you'd keep it as is... otherwise you add one to it. That gets you a result range of 0..9 without 4. Suppose we get 7 that way.

On the third iteration you'd generate a number in the range 0..7. If the generated number is less than 4, you'd keep it as is. If it's 4 or 5, you'd add one. If it's 6 or 7, you'd add two. That way the result range is 0..9 without 4 or 6.

Java - Generates random numbers and classify them for each value

Run the for loop 24 times instead of 25 times, since if i = 24, there is no array value at index randNum[24] (array indexes start at 0) (when you do i+1 inside the if statement).

for (int i = 0; i < 24; i++)
{
  //your if-else blocks
}

Java Generate Random Number Between Two Given Values

You could use e.g. r.nextInt(101)

For a more generic "in between two numbers" use:

Random r = new Random();
int low = 10;
int high = 100;
int result = r.nextInt(high-low) + low;

This gives you a random number in between 10 (inclusive) and 100 (exclusive)

Generating a Random Number between 1 and 10 Java

As the documentation says, this method call returns "a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive)". This means that you will get numbers from 0 to 9 in your case. So you've done everything correctly by adding one to that number.

Generally speaking, if you need to generate numbers from min to max (including both), you write

random.nextInt(max - min + 1) + min

Generating Unique Random Numbers in Java

• Add each number in the range sequentially in a list structure.
• Shuffle it.
• Take the first 'n'.

Here is a simple implementation. This will print 3 unique random numbers from the range 1-10.

import java.util.ArrayList;
import java.util.Collections;

public class UniqueRandomNumbers {
    
    public static void main(String[] args) {
        ArrayList<Integer> list = new ArrayList<Integer>();
        for (int i=1; i<11; i++) list.add(i);
        Collections.shuffle(list);
        for (int i=0; i<3; i++) System.out.println(list.get(i));
    }
}

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The first part of the fix with the original approach, as Mark Byers pointed out in an answer now deleted, is to use only a single Random instance.

That is what is causing the numbers to be identical. A Random instance is seeded by the current time in milliseconds. For a particular seed value, the 'random' instance will return the exact same sequence of pseudo random numbers.

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