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

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)
}

其他回答

我只是使用Math.random()生成一个随机数,然后将其乘以一个大数,比方说10000。因此,我得到一个介于0到10000之间的数字,并将其称为I。现在,如果我需要介于(x,y)之间的数字时,请执行以下操作:

i = x + (i % (y - x));

所以,所有的i都是x和y之间的数字。

要消除注释中指出的偏差,而不是将其乘以10000(或大数字),请将其乘以(y-x)。

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);
Random rng = new Random();
int min = 3;
int max = 11;
int upperBound = max - min + 1; // upper bound is exclusive, so +1
int num = min + rng.nextInt(upperBound);
System.out.println(num);
int randomNum = 5 + (int)(Math.random()*5);

范围5-10

如果掷骰子,它将是1到6(而不是0到6)之间的随机数,因此:

face = 1 + randomNumbers.nextInt(6);