在Java中生成10位唯一的随机数

Question

我正在尝试使用下面的代码来生成10位唯一的随机数。根据我的要求，我必须创建大约5000个唯一编号。这无法正常工作。它还会生成-
ve数字。同样，有时生成的数字中缺少一两位数字，导致8或9个数字而不是10。

public static synchronized  List generateRandomPin(){

    int START =1000000000;
    //int END = Integer.parseInt("9999999999");
    //long END = Integer.parseInt("9999999999");
    long END = 9999999999L;

    Random random = new Random();

    for (int idx = 1; idx <= 3000; ++idx){
        createRandomInteger(START, END, random);
    }

    return null;
}


private static void createRandomInteger(int aStart, long 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;
    logger.info("range>>>>>>>>>>>"+range);
    // compute a fraction of the range, 0 <= frac < range
    long fraction = (long)(range * aRandom.nextDouble());
    logger.info("fraction>>>>>>>>>>>>>>>>>>>>"+fraction);
    int randomNumber =  (int)(fraction + aStart);    
    logger.info("Generated : " + randomNumber);

  }

Answer 1

我认为您获得8/9位数字值和负数的原因是您要添加fraction一个long（带符号的64位值），该值可能大于的正数int范围（32位值）aStart。

该值正在溢出，randomNumber处于负32位范围内或几乎缠绕到aStart（由于int是带符号的32位值，您fraction只需要稍微小于（2
^ 32- aStart）就可以看到8或9位数字）。

您需要使用long所有值。

   private static void createRandomInteger(int aStart, long 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 = aEnd - (long)aStart + 1;
    logger.info("range>>>>>>>>>>>"+range);
    // compute a fraction of the range, 0 <= frac < range
    long fraction = (long)(range * aRandom.nextDouble());
    logger.info("fraction>>>>>>>>>>>>>>>>>>>>"+fraction);
    long randomNumber =  fraction + (long)aStart;    
    logger.info("Generated : " + randomNumber);

  }