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

下面是一个函数,它按照用户42155的请求,在lowerBoundIncluded和upperBoundIncluded定义的范围内返回一个整数随机数

SplitableRandom splitableRandom=新的Splitablerandom();

BiFunction<Integer,Integer,Integer> randomInt = (lowerBoundIncluded, upperBoundIncluded)
    -> splittableRandom.nextInt(lowerBoundIncluded, upperBoundIncluded + 1);

randomInt.apply(…,…);//获取随机数

…或更短,用于一次性生成随机数

new SplittableRandom().nextInt(lowerBoundIncluded, upperBoundIncluded + 1);

其他回答

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.

Use:

minValue + rn.nextInt(maxValue - minValue + 1)

只需对第一个解决方案稍作修改即可。

Random rand = new Random();
randomNum = minimum + rand.nextInt((maximum - minimum) + 1);

有关Random的实现,请参阅此处的更多信息

您可以使用以下代码段来解决问题:

Random r = new Random();
int myRandomNumber = 0;
myRandomNumber = r.nextInt(maxValue - minValue + 1) + minValue;

使用myRandomNumber(这将为您提供一个范围内的数字)。

我的一个朋友今天在大学里问过我同样的问题(他的要求是生成一个介于1和-1之间的随机数)。所以我写了这个,到目前为止,它在我的测试中运行良好。理想情况下,有很多方法可以在给定范围内生成随机数。试试看:

功能:

private static float getRandomNumberBetween(float numberOne, float numberTwo) throws Exception{

    if (numberOne == numberTwo){
        throw new Exception("Both the numbers can not be equal");
    }

    float rand = (float) Math.random();
    float highRange = Math.max(numberOne, numberTwo);
    float lowRange = Math.min(numberOne, numberTwo);

    float lowRand = (float) Math.floor(rand-1);
    float highRand = (float) Math.ceil(rand+1);

    float genRand = (highRange-lowRange)*((rand-lowRand)/(highRand-lowRand))+lowRange;

    return genRand;
}

执行方式如下:

System.out.println( getRandomNumberBetween(1,-1));