BigInteger：在可扩展的方法中计算小数位数

Question

我需要一个大整数的小数位数。例如：

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

我需要对biginger和184948十进制数字及更多数字执行此操作。我如何快速且可扩展地完成这项工作？

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10倍循环的方法更慢：

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

我认为可以使用bitLength（）获得log2值，然后将基数改为10。

然而，结果可能错了一位数，所以这只是一个近似值。

然而，如果这是可以接受的，你总是可以给结果加1，并将其限制为最多。或者，减去1，至少得到。

Answer 2

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 3

下面是一个基于Dariusz答案的快速方法：

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");
}

这可以在一秒钟内检查具有184,948十进制数字和更多数字的BigIntger。