一尘不染

表示Java分数的最佳方法?

java

我正在尝试使用Java中的分数。

我想实现算术函数。为此,我首先需要一种将功能标准化的方法。我知道我只有一个公分母才能将1/6和1/2加起来。我将不得不添加1/6和3/6。天真的方法会让我加2/12和6/12,然后减少。如何以最少的性能损失获得一个共同的分母?哪种算法最适合呢?

版本8(感谢hstoerr):

改进包括:

现在equals()方法与compareTo()方法一致

final class Fraction extends Number {
    private int numerator;
    private int denominator;

    public Fraction(int numerator, int denominator) {
        if(denominator == 0) {
            throw new IllegalArgumentException("denominator is zero");
        }
        if(denominator < 0) {
            numerator *= -1;
            denominator *= -1;
        }
        this.numerator = numerator;
        this.denominator = denominator;
    }

    public Fraction(int numerator) {
        this.numerator = numerator;
        this.denominator = 1;
    }

    public int getNumerator() {
        return this.numerator;
    }

    public int getDenominator() {
        return this.denominator;
    }

    public byte byteValue() {
        return (byte) this.doubleValue();
    }

    public double doubleValue() {
        return ((double) numerator)/((double) denominator);
    }

    public float floatValue() {
        return (float) this.doubleValue();
    }

    public int intValue() {
        return (int) this.doubleValue();
    }

    public long longValue() {
        return (long) this.doubleValue();
    }

    public short shortValue() {
        return (short) this.doubleValue();
    }

    public boolean equals(Fraction frac) {
        return this.compareTo(frac) == 0;
    }

    public int compareTo(Fraction frac) {
        long t = this.getNumerator() * frac.getDenominator();
        long f = frac.getNumerator() * this.getDenominator();
        int result = 0;
        if(t>f) {
            result = 1;
        }
        else if(f>t) {
            result = -1;
        }
        return result;
    }
}

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2020-03-19

共1个答案

一尘不染

碰巧的是不久前我写了一个BigFraction类,用于解决Euler项目问题。它保留了BigInteger分子和分母,因此它将永远不会溢出。但是,对于许多你永远不会溢出的操作来说,这会有点慢。无论如何,请根据需要使用它。我一直很想以某种方式炫耀它。:)

编辑:该代码的最新,最出色的版本,包括单元测试,现在托管在GitHub上,也可以通过Maven Central获得。我将原始代码留在这里,以便此答案不仅仅是一个链接…

import java.math.*;

/**
 * Arbitrary-precision fractions, utilizing BigIntegers for numerator and
 * denominator.  Fraction is always kept in lowest terms.  Fraction is
 * immutable, and guaranteed not to have a null numerator or denominator.
 * Denominator will always be positive (so sign is carried by numerator,
 * and a zero-denominator is impossible).
 */
public final class BigFraction extends Number implements Comparable<BigFraction>
{
  private static final long serialVersionUID = 1L; //because Number is Serializable
  private final BigInteger numerator;
  private final BigInteger denominator;

  public final static BigFraction ZERO = new BigFraction(BigInteger.ZERO, BigInteger.ONE, true);
  public final static BigFraction ONE = new BigFraction(BigInteger.ONE, BigInteger.ONE, true);

  /**
   * Constructs a BigFraction with given numerator and denominator.  Fraction
   * will be reduced to lowest terms.  If fraction is negative, negative sign will
   * be carried on numerator, regardless of how the values were passed in.
   */
  public BigFraction(BigInteger numerator, BigInteger denominator)
  {
    if(numerator == null)
      throw new IllegalArgumentException("Numerator is null");
    if(denominator == null)
      throw new IllegalArgumentException("Denominator is null");
    if(denominator.equals(BigInteger.ZERO))
      throw new ArithmeticException("Divide by zero.");

    //only numerator should be negative.
    if(denominator.signum() < 0)
    {
      numerator = numerator.negate();
      denominator = denominator.negate();
    }

    //create a reduced fraction
    BigInteger gcd = numerator.gcd(denominator);
    this.numerator = numerator.divide(gcd);
    this.denominator = denominator.divide(gcd);
  }

  /**
   * Constructs a BigFraction from a whole number.
   */
  public BigFraction(BigInteger numerator)
  {
    this(numerator, BigInteger.ONE, true);
  }

  public BigFraction(long numerator, long denominator)
  {
    this(BigInteger.valueOf(numerator), BigInteger.valueOf(denominator));
  }

  public BigFraction(long numerator)
  {
    this(BigInteger.valueOf(numerator), BigInteger.ONE, true);
  }

  /**
   * Constructs a BigFraction from a floating-point number.
   * 
   * Warning: round-off error in IEEE floating point numbers can result
   * in answers that are unexpected.  For example, 
   *     System.out.println(new BigFraction(1.1))
   * will print:
   *     2476979795053773/2251799813685248
   * 
   * This is because 1.1 cannot be expressed exactly in binary form.  The
   * given fraction is exactly equal to the internal representation of
   * the double-precision floating-point number.  (Which, for 1.1, is:
   * (-1)^0 * 2^0 * (1 + 0x199999999999aL / 0x10000000000000L).)
   * 
   * NOTE: In many cases, BigFraction(Double.toString(d)) may give a result
   * closer to what the user expects.
   */
  public BigFraction(double d)
  {
    if(Double.isInfinite(d))
      throw new IllegalArgumentException("double val is infinite");
    if(Double.isNaN(d))
      throw new IllegalArgumentException("double val is NaN");

    //special case - math below won't work right for 0.0 or -0.0
    if(d == 0)
    {
      numerator = BigInteger.ZERO;
      denominator = BigInteger.ONE;
      return;
    }

    final long bits = Double.doubleToLongBits(d);
    final int sign = (int)(bits >> 63) & 0x1;
    final int exponent = ((int)(bits >> 52) & 0x7ff) - 0x3ff;
    final long mantissa = bits & 0xfffffffffffffL;

    //number is (-1)^sign * 2^(exponent) * 1.mantissa
    BigInteger tmpNumerator = BigInteger.valueOf(sign==0 ? 1 : -1);
    BigInteger tmpDenominator = BigInteger.ONE;

    //use shortcut: 2^x == 1 << x.  if x is negative, shift the denominator
    if(exponent >= 0)
      tmpNumerator = tmpNumerator.multiply(BigInteger.ONE.shiftLeft(exponent));
    else
      tmpDenominator = tmpDenominator.multiply(BigInteger.ONE.shiftLeft(-exponent));

    //1.mantissa == 1 + mantissa/2^52 == (2^52 + mantissa)/2^52
    tmpDenominator = tmpDenominator.multiply(BigInteger.valueOf(0x10000000000000L));
    tmpNumerator = tmpNumerator.multiply(BigInteger.valueOf(0x10000000000000L + mantissa));

    BigInteger gcd = tmpNumerator.gcd(tmpDenominator);
    numerator = tmpNumerator.divide(gcd);
    denominator = tmpDenominator.divide(gcd);
  }

  /**
   * Constructs a BigFraction from two floating-point numbers.
   * 
   * Warning: round-off error in IEEE floating point numbers can result
   * in answers that are unexpected.  See BigFraction(double) for more
   * information.
   * 
   * NOTE: In many cases, BigFraction(Double.toString(numerator) + "/" + Double.toString(denominator))
   * may give a result closer to what the user expects.
   */
  public BigFraction(double numerator, double denominator)
  {
    if(denominator == 0)
      throw new ArithmeticException("Divide by zero.");

    BigFraction tmp = new BigFraction(numerator).divide(new BigFraction(denominator));
    this.numerator = tmp.numerator;
    this.denominator = tmp.denominator;
  }

  /**
   * Constructs a new BigFraction from the given BigDecimal object.
   */
  public BigFraction(BigDecimal d)
  {
    this(d.scale() < 0 ? d.unscaledValue().multiply(BigInteger.TEN.pow(-d.scale())) : d.unscaledValue(),
         d.scale() < 0 ? BigInteger.ONE                                             : BigInteger.TEN.pow(d.scale()));
  }

  public BigFraction(BigDecimal numerator, BigDecimal denominator)
  {
    if(denominator.equals(BigDecimal.ZERO))
      throw new ArithmeticException("Divide by zero.");

    BigFraction tmp = new BigFraction(numerator).divide(new BigFraction(denominator));
    this.numerator = tmp.numerator;
    this.denominator = tmp.denominator;
  }

  /**
   * Constructs a BigFraction from a String.  Expected format is numerator/denominator,
   * but /denominator part is optional.  Either numerator or denominator may be a floating-
   * point decimal number, which in the same format as a parameter to the
   * <code>BigDecimal(String)</code> constructor.
   * 
   * @throws NumberFormatException  if the string cannot be properly parsed.
   */
  public BigFraction(String s)
  {
    int slashPos = s.indexOf('/');
    if(slashPos < 0)
    {
      BigFraction res = new BigFraction(new BigDecimal(s));
      this.numerator = res.numerator;
      this.denominator = res.denominator;
    }
    else
    {
      BigDecimal num = new BigDecimal(s.substring(0, slashPos));
      BigDecimal den = new BigDecimal(s.substring(slashPos+1, s.length()));
      BigFraction res = new BigFraction(num, den);
      this.numerator = res.numerator;
      this.denominator = res.denominator;
    }
  }

  /**
   * Returns this + f.
   */
  public BigFraction add(BigFraction f)
  {
    if(f == null)
      throw new IllegalArgumentException("Null argument");

    //n1/d1 + n2/d2 = (n1*d2 + d1*n2)/(d1*d2) 
    return new BigFraction(numerator.multiply(f.denominator).add(denominator.multiply(f.numerator)),
                           denominator.multiply(f.denominator));
  }

  /**
   * Returns this + b.
   */
  public BigFraction add(BigInteger b)
  {
    if(b == null)
      throw new IllegalArgumentException("Null argument");

    //n1/d1 + n2 = (n1 + d1*n2)/d1
    return new BigFraction(numerator.add(denominator.multiply(b)),
                           denominator, true);
  }

  /**
   * Returns this + n.
   */
  public BigFraction add(long n)
  {
    return add(BigInteger.valueOf(n));
  }

  /**
   * Returns this - f.
   */
  public BigFraction subtract(BigFraction f)
  {
    if(f == null)
      throw new IllegalArgumentException("Null argument");

    return new BigFraction(numerator.multiply(f.denominator).subtract(denominator.multiply(f.numerator)),
                           denominator.multiply(f.denominator));
  }

  /**
   * Returns this - b.
   */
  public BigFraction subtract(BigInteger b)
  {
    if(b == null)
      throw new IllegalArgumentException("Null argument");

    return new BigFraction(numerator.subtract(denominator.multiply(b)),
                           denominator, true);
  }

  /**
   * Returns this - n.
   */
  public BigFraction subtract(long n)
  {
    return subtract(BigInteger.valueOf(n));
  }

  /**
   * Returns this * f.
   */
  public BigFraction multiply(BigFraction f)
  {
    if(f == null)
      throw new IllegalArgumentException("Null argument");

    return new BigFraction(numerator.multiply(f.numerator), denominator.multiply(f.denominator));
  }

  /**
   * Returns this * b.
   */
  public BigFraction multiply(BigInteger b)
  {
    if(b == null)
      throw new IllegalArgumentException("Null argument");

    return new BigFraction(numerator.multiply(b), denominator);
  }

  /**
   * Returns this * n.
   */
  public BigFraction multiply(long n)
  {
    return multiply(BigInteger.valueOf(n));
  }

  /**
   * Returns this / f.
   */
  public BigFraction divide(BigFraction f)
  {
    if(f == null)
      throw new IllegalArgumentException("Null argument");

    if(f.numerator.equals(BigInteger.ZERO))
      throw new ArithmeticException("Divide by zero");

    return new BigFraction(numerator.multiply(f.denominator), denominator.multiply(f.numerator));
  }

  /**
   * Returns this / b.
   */
  public BigFraction divide(BigInteger b)
  {
    if(b == null)
      throw new IllegalArgumentException("Null argument");

    if(b.equals(BigInteger.ZERO))
      throw new ArithmeticException("Divide by zero");

    return new BigFraction(numerator, denominator.multiply(b));
  }

  /**
   * Returns this / n.
   */
  public BigFraction divide(long n)
  {
    return divide(BigInteger.valueOf(n));
  }

  /**
   * Returns this^exponent.
   */
  public BigFraction pow(int exponent)
  {
    if(exponent == 0)
      return BigFraction.ONE;
    else if (exponent == 1)
      return this;
    else if (exponent < 0)
      return new BigFraction(denominator.pow(-exponent), numerator.pow(-exponent), true);
    else
      return new BigFraction(numerator.pow(exponent), denominator.pow(exponent), true);
  }

  /**
   * Returns 1/this.
   */
  public BigFraction reciprocal()
  {
    if(this.numerator.equals(BigInteger.ZERO))
      throw new ArithmeticException("Divide by zero");

    return new BigFraction(denominator, numerator, true);
  }

  /**
   * Returns the complement of this fraction, which is equal to 1 - this.
   * Useful for probabilities/statistics.

   */
  public BigFraction complement()
  {
    return new BigFraction(denominator.subtract(numerator), denominator, true);
  }

  /**
   * Returns -this.
   */
  public BigFraction negate()
  {
    return new BigFraction(numerator.negate(), denominator, true);
  }

  /**
   * Returns -1, 0, or 1, representing the sign of this fraction.
   */
  public int signum()
  {
    return numerator.signum();
  }

  /**
   * Returns the absolute value of this.
   */
  public BigFraction abs()
  {
    return (signum() < 0 ? negate() : this);
  }

  /**
   * Returns a string representation of this, in the form
   * numerator/denominator.
   */
  public String toString()
  {
    return numerator.toString() + "/" + denominator.toString();
  }

  /**
   * Returns if this object is equal to another object.
   */
  public boolean equals(Object o)
  {
    if(!(o instanceof BigFraction))
      return false;

    BigFraction f = (BigFraction)o;
    return numerator.equals(f.numerator) && denominator.equals(f.denominator);
  }

  /**
   * Returns a hash code for this object.
   */
  public int hashCode()
  {
    //using the method generated by Eclipse, but streamlined a bit..
    return (31 + numerator.hashCode())*31 + denominator.hashCode();
  }

  /**
   * Returns a negative, zero, or positive number, indicating if this object
   * is less than, equal to, or greater than f, respectively.
   */
  public int compareTo(BigFraction f)
  {
    if(f == null)
      throw new IllegalArgumentException("Null argument");

    //easy case: this and f have different signs
    if(signum() != f.signum())
      return signum() - f.signum();

    //next easy case: this and f have the same denominator
    if(denominator.equals(f.denominator))
      return numerator.compareTo(f.numerator);

    //not an easy case, so first make the denominators equal then compare the numerators 
    return numerator.multiply(f.denominator).compareTo(denominator.multiply(f.numerator));
  }

  /**
   * Returns the smaller of this and f.
   */
  public BigFraction min(BigFraction f)
  {
    if(f == null)
      throw new IllegalArgumentException("Null argument");

    return (this.compareTo(f) <= 0 ? this : f);
  }

  /**
   * Returns the maximum of this and f.
   */
  public BigFraction max(BigFraction f)
  {
    if(f == null)
      throw new IllegalArgumentException("Null argument");

    return (this.compareTo(f) >= 0 ? this : f);
  }

  /**
   * Returns a positive BigFraction, greater than or equal to zero, and less than one.
   */
  public static BigFraction random()
  {
    return new BigFraction(Math.random());
  }

  public final BigInteger getNumerator() { return numerator; }
  public final BigInteger getDenominator() { return denominator; }

  //implementation of Number class.  may cause overflow.
  public byte   byteValue()   { return (byte) Math.max(Byte.MIN_VALUE,    Math.min(Byte.MAX_VALUE,    longValue())); }
  public short  shortValue()  { return (short)Math.max(Short.MIN_VALUE,   Math.min(Short.MAX_VALUE,   longValue())); }
  public int    intValue()    { return (int)  Math.max(Integer.MIN_VALUE, Math.min(Integer.MAX_VALUE, longValue())); }
  public long   longValue()   { return Math.round(doubleValue()); }
  public float  floatValue()  { return (float)doubleValue(); }
  public double doubleValue() { return toBigDecimal(18).doubleValue(); }

  /**
   * Returns a BigDecimal representation of this fraction.  If possible, the
   * returned value will be exactly equal to the fraction.  If not, the BigDecimal
   * will have a scale large enough to hold the same number of significant figures
   * as both numerator and denominator, or the equivalent of a double-precision
   * number, whichever is more.
   */
  public BigDecimal toBigDecimal()
  {
    //Implementation note:  A fraction can be represented exactly in base-10 iff its
    //denominator is of the form 2^a * 5^b, where a and b are nonnegative integers.
    //(In other words, if there are no prime factors of the denominator except for
    //2 and 5, or if the denominator is 1).  So to determine if this denominator is
    //of this form, continually divide by 2 to get the number of 2's, and then
    //continually divide by 5 to get the number of 5's.  Afterward, if the denominator
    //is 1 then there are no other prime factors.

    //Note: number of 2's is given by the number of trailing 0 bits in the number
    int twos = denominator.getLowestSetBit();
    BigInteger tmpDen = denominator.shiftRight(twos); // x / 2^n === x >> n

    final BigInteger FIVE = BigInteger.valueOf(5);
    int fives = 0;
    BigInteger[] divMod = null;

    //while(tmpDen % 5 == 0) { fives++; tmpDen /= 5; }
    while(BigInteger.ZERO.equals((divMod = tmpDen.divideAndRemainder(FIVE))[1]))
    {
      fives++;
      tmpDen = divMod[0];
    }

    if(BigInteger.ONE.equals(tmpDen))
    {
      //This fraction will terminate in base 10, so it can be represented exactly as
      //a BigDecimal.  We would now like to make the fraction of the form
      //unscaled / 10^scale.  We know that 2^x * 5^x = 10^x, and our denominator is
      //in the form 2^twos * 5^fives.  So use max(twos, fives) as the scale, and
      //multiply the numerator and deminator by the appropriate number of 2's or 5's
      //such that the denominator is of the form 2^scale * 5^scale.  (Of course, we
      //only have to actually multiply the numerator, since all we need for the
      //BigDecimal constructor is the scale.
      BigInteger unscaled = numerator;
      int scale = Math.max(twos, fives);

      if(twos < fives)
        unscaled = unscaled.shiftLeft(fives - twos); //x * 2^n === x << n
      else if (fives < twos)
        unscaled = unscaled.multiply(FIVE.pow(twos - fives));

      return new BigDecimal(unscaled, scale);
    }

    //else: this number will repeat infinitely in base-10.  So try to figure out
    //a good number of significant digits.  Start with the number of digits required
    //to represent the numerator and denominator in base-10, which is given by
    //bitLength / log[2](10).  (bitLenth is the number of digits in base-2).
    final double LG10 = 3.321928094887362; //Precomputed ln(10)/ln(2), a.k.a. log[2](10)
    int precision = Math.max(numerator.bitLength(), denominator.bitLength());
    precision = (int)Math.ceil(precision / LG10);

    //If the precision is less than 18 digits, use 18 digits so that the number
    //will be at least as accurate as a cast to a double.  For example, with
    //the fraction 1/3, precision will be 1, giving a result of 0.3.  This is
    //quite a bit different from what a user would expect.
    if(precision < 18)
      precision = 18;

    return toBigDecimal(precision);
  }

  /**
   * Returns a BigDecimal representation of this fraction, with a given precision.
   * @param precision  the number of significant figures to be used in the result.
   */
  public BigDecimal toBigDecimal(int precision)
  {
    return new BigDecimal(numerator).divide(new BigDecimal(denominator), new MathContext(precision, RoundingMode.HALF_EVEN));
  }

  //--------------------------------------------------------------------------
  //  PRIVATE FUNCTIONS
  //--------------------------------------------------------------------------

  /**
   * Private constructor, used when you can be certain that the fraction is already in
   * lowest terms.  No check is done to reduce numerator/denominator.  A check is still
   * done to maintain a positive denominator.
   * 
   * @param throwaway  unused variable, only here to signal to the compiler that this
   *                   constructor should be used.
   */
  private BigFraction(BigInteger numerator, BigInteger denominator, boolean throwaway)
  {
    if(denominator.signum() < 0)
    {
      this.numerator = numerator.negate();
      this.denominator = denominator.negate();
    }
    else
    {
      this.numerator = numerator;
      this.denominator = denominator;
    }
  }

}
2020-03-19