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

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);
    }
}

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。

在java-8中,他们在Random类中引入了方法int(int randomNumberOrigin,int randomNumber Bound)。

例如,如果要生成[0,10]范围内的五个随机整数(或单个整数),只需执行以下操作:

Random r = new Random();
int[] fiveRandomNumbers = r.ints(5, 0, 11).toArray();
int randomNumber = r.ints(1, 0, 11).findFirst().getAsInt();

第一个参数仅指示生成的IntStream的大小(这是生成无限IntStream的重载方法)。

如果需要执行多个单独的调用,可以从流中创建无限基元迭代器:

public final class IntRandomNumberGenerator {

    private PrimitiveIterator.OfInt randomIterator;

    /**
     * Initialize a new random number generator that generates
     * random numbers in the range [min, max]
     * @param min - the min value (inclusive)
     * @param max - the max value (inclusive)
     */
    public IntRandomNumberGenerator(int min, int max) {
        randomIterator = new Random().ints(min, max + 1).iterator();
    }

    /**
     * Returns a random number in the range (min, max)
     * @return a random number in the range (min, max)
     */
    public int nextInt() {
        return randomIterator.nextInt();
    }
}

您也可以对双值和长值执行此操作。

你可以使用

RandomStringUtils.randomNumeric(int count)

该方法也来自Apache Commons。

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:

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。