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

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。

其他回答

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);
    }
}
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);

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。

使用nexttint(n)方法为最小值和最大值的差值生成一个随机数,然后将最小值添加到结果中:

Random rn = new Random();
int result = rn.nextInt(max - min + 1) + min;
System.out.println(result);

如果你不想重新发明轮子,这里有一个简单明了的解决方案:

RandomStringUtils.randomNumeric(count); // Where count is the number of digits you want in the random number.

Apache Commons,例如组件Lang中的RandomUtils,提供了许多选项来生成任意格式的随机数。