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.

其他回答

您可以按以下方式操作。

import java.util.Random;
public class RandomTestClass {

    public static void main(String[] args) {
        Random r = new Random();
        int max, min;
        Scanner scanner = new Scanner(System.in);
        System.out.println("Enter maximum value : ");
        max = scanner.nextInt();
        System.out.println("Enter minimum value : ");
        min = scanner.nextInt();
        int randomNum;
        randomNum = r.nextInt(max) + min;
        System.out.println("Random Number : " + randomNum);
    }

}

下面是一个简单的示例,它显示了如何从闭合的[min,max]范围生成随机数,而min<=max为真

您可以将其作为孔中字段类重用,也可以将所有Random.class方法放在一个位置

结果示例:

RandomUtils random = new RandomUtils();
random.nextInt(0, 0); // returns 0
random.nextInt(10, 10); // returns 10
random.nextInt(-10, 10); // returns numbers from -10 to 10 (-10, -9....9, 10)
random.nextInt(10, -10); // throws assert

来源:

import junit.framework.Assert;
import java.util.Random;

public class RandomUtils extends Random {

    /**
     * @param min generated value. Can't be > then max
     * @param max generated value
     * @return values in closed range [min, max].
     */
    public int nextInt(int min, int max) {
        Assert.assertFalse("min can't be > then max; values:[" + min + ", " + max + "]", min > max);
        if (min == max) {
            return max;
        }

        return nextInt(max - min + 1) + min;
    }
}

Java中的Math.Random类是基于0的。所以,如果你这样写:

Random rand = new Random();
int x = rand.nextInt(10);

x将介于0-9之间(含0-9)。

因此,给定以下25项的数组,生成0(数组的基数)和array.length之间的随机数的代码为:

String[] i = new String[25];
Random rand = new Random();
int index = 0;

index = rand.nextInt( i.length );

由于i.length将返回25,因此nextInt(i.length)将返回0-24之间的数字。另一个选项是Math.Random,其工作方式相同。

index = (int) Math.floor(Math.random() * i.length);

为了更好地理解,请查看论坛帖子Random Intervals(archive.org)。

private static Random random = new Random();    

public static int getRandomInt(int min, int max){
  return random.nextInt(max - min + 1) + min;
}

OR

public static int getRandomInt(Random random, int min, int max)
{
  return random.nextInt(max - min + 1) + min;
}

我想知道Apache Commons Math库提供的任何随机数生成方法是否符合要求。

例如:RandomDataGenerator.nextInt或RandomDataGenerator.nextLong