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

可以使用以下代码:

ThreadLocalRandom.current().nextInt(rangeStart, rangeEndExclusive)

其他回答

您可以使用Random类生成随机数,然后使用.nextInt(maxNumber)生成随机数。maxNumber是生成随机数时所需的最大值。请记住,Random类给你的是0到maxNumber-1的数字。

Random r = new Random();
int i = r.nextInt();

另一种方法是使用Math.Random()类,许多学校的课程都要求您使用该类,因为它更高效,而且您不必声明新的Random对象。要使用Math.random()获取随机数,请键入:

Math.random() * (max - min) + min;

在A和b之间生成n个随机数的简单方法例如a=90,b=100,n=20

Random r = new Random();
for(int i =0; i<20; i++){
    System.out.println(r.ints(90, 100).iterator().nextInt());
}

r.ints()返回一个IntStream,并有几个有用的方法,看看它的API。

我认为这段代码会奏效。请尝试一下:

import java.util.Random;
public final class RandomNumber {

    public static final void main(String... aArgs) {
        log("Generating 10 random integers in range 1..10.");
        int START = 1;
        int END = 10;
        Random randomGenerator = new Random();
        for (int idx=1; idx<=10; ++idx) {

            // int randomInt=randomGenerator.nextInt(100);
            // log("Generated : " + randomInt);
            showRandomInteger(START,END,randomGenerator);
        }
        log("Done");
    }

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

    private static void showRandomInteger(int aStart, int aEnd, Random aRandom) {
        if (aStart > aEnd) {
            throw new IllegalArgumentException("Start cannot exceed End.");
        }
        long range = (long)aEnd - (long)aStart + 1;
        long fraction = (long) (range * aRandom.nextDouble());
        int randomNumber = (int) (fraction + aStart);
        log("Generated" + randomNumber);
    }
}

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;
    }
}

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.