Generating a random number between multiple ranges

问题

I understand how to make a random number which is between two numbers:

1 + (int)(Math.random() * ((10 - 1) + 1))

or

Min + (int)(Math.random() * ((Max - Min) + 1))

But how do I go about generating a random number which falls into multiple ranges?

For example: number can be between 1 to 10 or between 50 to 60

回答1:

I'd go with something like this, to allow you to do it with as many ranges as you like:

import java.util.ArrayList;
import java.util.List;
import java.util.Random;

class RandomInRanges
{
    private final List<Integer> range = new ArrayList<>();

    RandomInRanges(int min, int max)
    {
        this.addRange(min, max);
    }

    final void addRange(int min, int max)
    {
        for(int i = min; i <= max; i++)
        {
            this.range.add(i);
        }
    }

    int getRandom()
    {
        return this.range.get(new Random().nextInt(this.range.size()));
    }

    public static void main(String[] args)
    {
        RandomInRanges rir = new RandomInRanges(1, 10);
        rir.addRange(50, 60);
        System.out.println(rir.getRandom());
    }
}

回答2:

First generate an integer between 1 and 20. Then if the value is above 10, map to the second interval.

Random random = new Random();

for (int i=0;i<100;i++) {
    int r = 1 + random.nextInt(60-50+10-1);
    if (r>10) r+=(50-10);
    System.out.println(r);      
}

回答3:

First, you need to know how many numbers are in each range. (I'm assuming you are choosing integers from a discrete range, not real values from a continuous range.) In your example, there are 10 integers in the first range, and 11 in the second. This means that 10/21 times, you should choose from the first range, and 11/21 times choose from the second. In pseudo-code

x = random(1,21)
if x in 1..10
   return random(1,10)
else
   return random(50,60)

回答4:

if the list is known I think you can use something like this.

public class Test
{
    public static void main(String[] args)
    {
        int a;
       for(int i=0;i<10;i++)
      {
            a=(int) (Math.random()*((10-0)+(60-50)));

            if(a<=10)
            {

            }
            else if(a>(60-50))
            {
                a=a+50;
        }
            System.out.println(a);
       }
    }
}

回答5:

How about the following approach: randomize picking a range an use that range to generage your random number, for that you'll have to put your ranges in some list or array

public class RandomRangeGenerator {

    class Range {
        public int min, max;
        public Range(int min, int max) { this.min = min; this.max = max; }
    }

    @Test
    public void generate() {
        List<Range> ranges = new ArrayList<>();
        ranges.add(new Range(1, 10));
        ranges.add(new Range(50, 60));
        List<Integer> randomNumbers = generateRandomNumbers(ranges, 10);
        System.out.println(randomNumbers.toString());
        for(Integer i : randomNumbers) {
            Assert.assertTrue((i >= 1 && i <= 10) || (i >= 50 && i <= 60));
        }
    }

    private List<Integer> generateRandomNumbers(List<Range> ranges, int numberOfNumbers) {
        List<Integer> randomNumbers = new ArrayList<>(numberOfNumbers+1);
        while(numberOfNumbers-- > 0) {
            Range range = ranges.get(new Random().nextInt(ranges.size()));
            randomNumbers.add(range.min + (int)(Math.random() * ((range.max - range.min) + 1)));
        }
        return randomNumbers;
    }
}

来源：https://stackoverflow.com/questions/15591173/generating-a-random-number-between-multiple-ranges