如何在特定范围内生成随机int值?

以下方法存在与整数溢出相关的错误:

randomNum = minimum + (int)(Math.random() * maximum);
// Bug: `randomNum` can be bigger than `maximum`.
Random rn = new Random();
int n = maximum - minimum + 1;
int i = rn.nextInt() % n;
randomNum =  minimum + i;
// Bug: `randomNum` can be smaller than `minimum`.

当前回答

我使用这个:

 /**
   * @param min - The minimum.
   * @param max - The maximum.
   * @return A random double between these numbers (inclusive the minimum and maximum).
   */
 public static double getRandom(double min, double max) {
   return (Math.random() * (max + 1 - min)) + min;
 }

如果需要,可以将其转换为整数。

其他回答

这里有一个有用的类,可以在包含/排除边界的任意组合范围内生成随机整数:

import java.util.Random;

public class RandomRange extends Random {
    public int nextIncInc(int min, int max) {
        return nextInt(max - min + 1) + min;
    }

    public int nextExcInc(int min, int max) {
        return nextInt(max - min) + 1 + min;
    }

    public int nextExcExc(int min, int max) {
        return nextInt(max - min - 1) + 1 + min;
    }

    public int nextIncExc(int min, int max) {
        return nextInt(max - min) + min;
    }
}

使用Java 8 IntStream和Collections.shuffle的不同方法

import java.util.stream.IntStream;
import java.util.ArrayList;
import java.util.Collections;

public class Main {

    public static void main(String[] args) {

        IntStream range = IntStream.rangeClosed(5,10);
        ArrayList<Integer> ls =  new ArrayList<Integer>();

        //populate the ArrayList
        range.forEach(i -> ls.add(new Integer(i)) );

        //perform a random shuffle  using the Collections Fisher-Yates shuffle
        Collections.shuffle(ls);
        System.out.println(ls);
    }
}

Scala中的等价项

import scala.util.Random

object RandomRange extends App{
  val x =  Random.shuffle(5 to 10)
    println(x)
}

尝试使用org.apachemons.lang.RandomStringUtils类。是的,它有时会给出一个相邻的重复数字,但它会给出介于5和15之间的值:

    while (true)
    {
        int abc = Integer.valueOf(RandomStringUtils.randomNumeric(1));
        int cd = Integer.valueOf(RandomStringUtils.randomNumeric(2));
        if ((cd-abc) >= 5 && (cd-abc) <= 15)
        {
            System.out.println(cd-abc);
            break;
        }
    }

我发现这个例子生成随机数:


此示例生成特定范围内的随机整数。

import java.util.Random;

/** Generate random integers in a certain range. */
public final class RandomRange {

  public static final void main(String... aArgs){
    log("Generating random integers in the range 1..10.");

    int START = 1;
    int END = 10;
    Random random = new Random();
    for (int idx = 1; idx <= 10; ++idx){
      showRandomInteger(START, END, random);
    }

    log("Done.");
  }

  private static void showRandomInteger(int aStart, int aEnd, Random aRandom){
    if ( aStart > aEnd ) {
      throw new IllegalArgumentException("Start cannot exceed End.");
    }
    //get the range, casting to long to avoid overflow problems
    long range = (long)aEnd - (long)aStart + 1;
    // compute a fraction of the range, 0 <= frac < range
    long fraction = (long)(range * aRandom.nextDouble());
    int randomNumber =  (int)(fraction + aStart);    
    log("Generated : " + randomNumber);
  }

  private static void log(String aMessage){
    System.out.println(aMessage);
  }
} 

此类的示例运行:

Generating random integers in the range 1..10.
Generated : 9
Generated : 3
Generated : 3
Generated : 9
Generated : 4
Generated : 1
Generated : 3
Generated : 9
Generated : 10
Generated : 10
Done.

在java-8中,他们在Random类中引入了方法int(int randomNumberOrigin,int randomNumber Bound)。

例如,如果要生成[0,10]范围内的五个随机整数(或单个整数),只需执行以下操作:

Random r = new Random();
int[] fiveRandomNumbers = r.ints(5, 0, 11).toArray();
int randomNumber = r.ints(1, 0, 11).findFirst().getAsInt();

第一个参数仅指示生成的IntStream的大小(这是生成无限IntStream的重载方法)。

如果需要执行多个单独的调用,可以从流中创建无限基元迭代器:

public final class IntRandomNumberGenerator {

    private PrimitiveIterator.OfInt randomIterator;

    /**
     * Initialize a new random number generator that generates
     * random numbers in the range [min, max]
     * @param min - the min value (inclusive)
     * @param max - the max value (inclusive)
     */
    public IntRandomNumberGenerator(int min, int max) {
        randomIterator = new Random().ints(min, max + 1).iterator();
    }

    /**
     * Returns a random number in the range (min, max)
     * @return a random number in the range (min, max)
     */
    public int nextInt() {
        return randomIterator.nextInt();
    }
}

您也可以对双值和长值执行此操作。