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

您可以编辑第二个代码示例以:

Random rn = new Random();
int range = maximum - minimum + 1;
int randomNum =  rn.nextInt(range) + minimum;

其他回答

请原谅我过于挑剔,但大多数人建议的解决方案,即min+rng.nextInt(max-min+1),似乎很危险,因为:

rng.nextInt(n)无法达到整数.MAX_VALUE。当min为负值时,(max-min)可能会导致溢出。

万无一失的解决方案将为[Integer.min_VALUE,Integer.max_VALUE]内的任何min<=max返回正确的结果。请考虑以下简单的实现:

int nextIntInRange(int min, int max, Random rng) {
   if (min > max) {
      throw new IllegalArgumentException("Cannot draw random int from invalid range [" + min + ", " + max + "].");
   }
   int diff = max - min;
   if (diff >= 0 && diff != Integer.MAX_VALUE) {
      return (min + rng.nextInt(diff + 1));
   }
   int i;
   do {
      i = rng.nextInt();
   } while (i < min || i > max);
   return i;
}

尽管效率低下,但请注意while循环中成功的概率始终为50%或更高。

Use:

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

让我们举个例子。

假设我希望生成5-10之间的数字:

int max = 10;
int min = 5;
int diff = max - min;
Random rn = new Random();
int i = rn.nextInt(diff + 1);
i += min;
System.out.print("The Random Number is " + i);

让我们了解这一点。。。

用最高值初始化max,用最低值初始化min。现在,我们需要确定可以获得多少可能的值。在本例中,应为:5, 6, 7, 8, 9, 10所以,这个计数应该是max-min+1。即10-5+1=6随机数将生成0-5之间的数字。即0、1、2、3、4、5将最小值添加到随机数将产生:5, 6, 7, 8, 9, 10 因此,我们获得了所需的范围。

只需执行以下语句即可完成:

Randomizer.generate(0, 10); // Minimum of zero and maximum of ten

下面是它的源代码。

文件Randomizer.java

public class Randomizer {
    public static int generate(int min, int max) {
        return min + (int)(Math.random() * ((max - min) + 1));
    }
}

它只是干净和简单。

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。