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

Use:

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

在Java 1.7或更高版本中,执行此操作的标准方法如下:

import java.util.concurrent.ThreadLocalRandom;

// nextInt is normally exclusive of the top value,
// so add 1 to make it inclusive
int randomNum = ThreadLocalRandom.current().nextInt(min, max + 1);

请参阅相关的JavaDoc。这种方法的优点是不需要显式初始化java.util.Random实例,如果使用不当,可能会导致混淆和错误。

然而,相反,没有办法明确设置种子,因此在测试或保存游戏状态等有用的情况下,很难再现结果。在这些情况下,可以使用下面所示的Java 1.7之前的技术。

在Java 1.7之前,执行此操作的标准方法如下:

import java.util.Random;

/**
 * Returns a pseudo-random number between min and max, inclusive.
 * The difference between min and max can be at most
 * <code>Integer.MAX_VALUE - 1</code>.
 *
 * @param min Minimum value
 * @param max Maximum value.  Must be greater than min.
 * @return Integer between min and max, inclusive.
 * @see java.util.Random#nextInt(int)
 */
public static int randInt(int min, int max) {

    // NOTE: This will (intentionally) not run as written so that folks
    // copy-pasting have to think about how to initialize their
    // Random instance.  Initialization of the Random instance is outside
    // the main scope of the question, but some decent options are to have
    // a field that is initialized once and then re-used as needed or to
    // use ThreadLocalRandom (if using at least Java 1.7).
    // 
    // In particular, do NOT do 'Random rand = new Random()' here or you
    // will get not very good / not very random results.
    Random rand;

    // nextInt is normally exclusive of the top value,
    // so add 1 to make it inclusive
    int randomNum = rand.nextInt((max - min) + 1) + min;

    return randomNum;
}

请参阅相关的JavaDoc。实际上,java.util.Random类通常比java.lang.Math.Random()更好。

特别是,当标准库中有一个简单的API来完成任务时,无需重新发明随机整数生成轮。

使用这些方法可能很方便:

此方法将返回提供的最小值和最大值之间的随机数:

public static int getRandomNumberBetween(int min, int max) {
    Random foo = new Random();
    int randomNumber = foo.nextInt(max - min) + min;
    if (randomNumber == min) {
        // Since the random number is between the min and max values, simply add 1
        return min + 1;
    } else {
        return randomNumber;
    }
}

并且该方法将从所提供的最小值和最大值返回随机数(因此生成的数也可以是最小值或最大值):

public static int getRandomNumberFrom(int min, int max) {
    Random foo = new Random();
    int randomNumber = foo.nextInt((max + 1) - min) + min;

    return randomNumber;
}

Use:

Random ran = new Random();
int x = ran.nextInt(6) + 5;

整数x现在是可能结果为5-10的随机数。

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));