java BigInteger：计算可伸缩方法中的小数位数

Question

我需要一个BigInteger的十进制数字计数。例如：

• 99返回2
• 1234返回4
• 9999返回4
• 12345678901234567890返回20

我需要对一个BigInteger有184948个十进制数字和更多的的BigInteger执行此操作如何快速且可扩展地执行此操作

转换为字符串的方法很慢：

public String getWritableNumber(BigInteger number) {
   // Takes over 30 seconds for 184948 decimal digits
   return "10^" + (number.toString().length() - 1);
}

这种十次循环的方法甚至更慢：

public String getWritableNumber(BigInteger number) {
    int digitSize = 0;
    while (!number.equals(BigInteger.ZERO)) {
        number = number.divide(BigInteger.TEN);
        digitSize++;
    }
    return "10^" + (digitSize - 1);
}

有没有更快的方法

Answer 1

这是比转换为字符串方法更快的另一种方法。虽然不是最佳运行时间，但仍然是合理的0.65秒，而使用Convert to String方法时为2.46秒（180000位）

此方法根据给定值计算以10为底对数的整数部分。然而，它没有使用循环除法，而是使用了一种类似于平方求幂的技术

下面是实现前面提到的运行时的粗略实现：

public static BigInteger log(BigInteger base,BigInteger num)
{
    /* The technique tries to get the products among the squares of base
     * close to the actual value as much as possible without exceeding it.
     * */
    BigInteger resultSet = BigInteger.ZERO;
    BigInteger actMult = BigInteger.ONE;
    BigInteger lastMult = BigInteger.ONE;
    BigInteger actor = base;
    BigInteger incrementor = BigInteger.ONE;
    while(actMult.multiply(base).compareTo(num)<1)
    {
        int count = 0;
        while(actMult.multiply(actor).compareTo(num)<1)
        {
            lastMult = actor; //Keep the old squares
            actor = actor.multiply(actor); //Square the base repeatedly until the value exceeds 
            if(count>0) incrementor = incrementor.multiply(BigInteger.valueOf(2));
            //Update the current exponent of the base
            count++;
        }
        if(count == 0) break;

        /* If there is no way to multiply the "actMult"
         * with squares of the base (including the base itself)
         * without keeping it below the actual value, 
         * it is the end of the computation 
         */
        actMult = actMult.multiply(lastMult);
        resultSet = resultSet.add(incrementor);
        /* Update the product and the exponent
         * */
        actor = base;
        incrementor = BigInteger.ONE;
        //Reset the values for another iteration
    }
    return resultSet;
}
public static int digits(BigInteger num)
{
    if(num.equals(BigInteger.ZERO)) return 1;
    if(num.compareTo(BigInteger.ZERO)<0) num = num.multiply(BigInteger.valueOf(-1));
    return log(BigInteger.valueOf(10),num).intValue()+1;
}

希望这会有所帮助

Answer 2

下面是一个基于Dariusz's answer的快速方法：

public static int getDigitCount(BigInteger number) {
  double factor = Math.log(2) / Math.log(10);
  int digitCount = (int) (factor * number.bitLength() + 1);
  if (BigInteger.TEN.pow(digitCount - 1).compareTo(number) > 0) {
    return digitCount - 1;
  }
  return digitCount;
}

以下代码测试数字1、9、10、99、100、999、1000等，一直测试到一万位：

public static void test() {
  for (int i = 0; i < 10000; i++) {
    BigInteger n = BigInteger.TEN.pow(i);
    if (getDigitCount(n.subtract(BigInteger.ONE)) != i || getDigitCount(n) != i + 1) {
      System.out.println("Failure: " + i);
    }
  }
  System.out.println("Done");
}

这可以检查BigInteger和184,948十进制数字，以及一秒以内的更多数字

Answer 3

您可以首先将BigInteger转换为BigDecimal，然后使用这个answer来计算位数。这似乎比使用BigInteger.toString()更有效，因为这将为String表示分配内存

    private static int numberOfDigits(BigInteger value) {
        return significantDigits(new BigDecimal(value));
    }

    private static int significantDigits(BigDecimal value) {
        return value.scale() < 0
               ? value.precision() - value.scale()
               : value.precision();
    }

Answer 4

这看起来很有效。我还没有运行详尽的测试，也没有运行任何时间测试，但它似乎有一个合理的运行时间

public class Test {
  /**
   * Optimised for huge numbers.
   *
   * http://en.wikipedia.org/wiki/Logarithm#Change_of_base
   *
   * States that log[b](x) = log[k](x)/log[k](b)
   *
   * We can get log[2](x) as the bitCount of the number so what we need is
   * essentially bitCount/log[2](10). Sadly that will lead to inaccuracies so
   * here I will attempt an iterative process that should achieve accuracy.
   *
   * log[2](10) = 3.32192809488736234787 so if I divide by 10^(bitCount/4) we
   * should not go too far. In fact repeating that process while adding (bitCount/4)
   * to the running count of the digits will end up with an accurate figure
   * given some twiddling at the end.
   * 
   * So here's the scheme:
   * 
   * While there are more than 4 bits in the number
   *   Divide by 10^(bits/4)
   *   Increase digit count by (bits/4)
   * 
   * Fiddle around to accommodate the remaining digit - if there is one.
   * 
   * Essentially - each time around the loop we remove a number of decimal 
   * digits (by dividing by 10^n) keeping a count of how many we've removed.
   * 
   * The number of digits we remove is estimated from the number of bits in the 
   * number (i.e. log[2](x) / 4). The perfect figure for the reduction would be
   * log[2](x) / 3.3219... so dividing by 4 is a good under-estimate. We 
   * don't go too far but it does mean we have to repeat it just a few times.
   */
  private int log10(BigInteger huge) {
    int digits = 0;
    int bits = huge.bitLength();
    // Serious reductions.
    while (bits > 4) {
      // 4 > log[2](10) so we should not reduce it too far.
      int reduce = bits / 4;
      // Divide by 10^reduce
      huge = huge.divide(BigInteger.TEN.pow(reduce));
      // Removed that many decimal digits.
      digits += reduce;
      // Recalculate bitLength
      bits = huge.bitLength();
    }
    // Now 4 bits or less - add 1 if necessary.
    if ( huge.intValue() > 9 ) {
      digits += 1;
    }
    return digits;
  }

  // Random tests.
  Random rnd = new Random();
  // Limit the bit length.
  int maxBits = BigInteger.TEN.pow(200000).bitLength();

  public void test() {
    // 100 tests.
    for (int i = 1; i <= 100; i++) {
      BigInteger huge = new BigInteger((int)(Math.random() * maxBits), rnd);
      // Note start time.
      long start = System.currentTimeMillis();
      // Do my method.
      int myLength = log10(huge);
      // Record my result.
      System.out.println("Digits: " + myLength+ " Took: " + (System.currentTimeMillis() - start));
      // Check the result.
      int trueLength = huge.toString().length() - 1;
      if (trueLength != myLength) {
        System.out.println("WRONG!! " + (myLength - trueLength));
      }
    }
  }

  public static void main(String args[]) {
    new Test().test();
  }

}

在我的Celeron M笔记本电脑上花了大约3秒，所以在一些像样的装备上应该会达到2秒以下

Answer 5

我认为可以使用bitLength()获得log2值，然后使用change the base to 10

结果可能是错误的，但是，一个数字，所以这只是一个近似值

但是，如果这是可以接受的，您可以始终将1添加到结果中，并将其绑定为最多。或者，减去1，得到至少