问题
My task is to develop a rational class. If 500 and 1000 are my inputs, then (½) must be my output. I have written a program on my own to find it.
Is there another best way to find the solution, or my program is already the best one?
public class Rational {
public static void main(String[] args){
int n1 = Integer.parseInt(args[0]);
int n2 = Integer.parseInt(args[1]);
int temp1 = n1;
int temp2 = n2;
while (n1 != n2){
if(n1 > n2)
n1 = n1 - n2;
else
n2 = n2 - n1;
}
int n3 = temp1 / n1 ;
int n4 = temp2 / n1 ;
System.out.print("\n Output :\n");
System.out.print(n3 + "/" + n4 + "\n\n" );
System.exit(0);
}
}
回答1:
Interesting question. Here's some executable code that does it in just a couple of lines:
/** @return the greatest common denominator */
public static long gcm(long a, long b) {
return b == 0 ? a : gcm(b, a % b); // Not bad for one line of code :)
}
public static String asFraction(long a, long b) {
long gcm = gcm(a, b);
return (a / gcm) + "/" + (b / gcm);
}
public static void main(String[] args) {
System.out.println(asFraction(500, 1000)); // "1/2"
System.out.println(asFraction(17, 3)); // "17/3"
System.out.println(asFraction(462, 1071)); // "22/51"
}
回答2:
You need the GCD. Either use BigInteger like Nathan mentioned or if you can't, use your own.
public int GCD(int a, int b){
if (b==0) return a;
return GCD(b,a%b);
}
Then you can divide each number by the GCD, like you have done above.
This will give you an improper fraction. If you need a mixed fraction then you can get the new numbers. Example if you had 1500 and 500 for inputs you would end up with 3/2 as your answer. Maybe you want 1 1/2. So you just divide 3/2 and get 1 and then get the remainder of 3/2 which is also 1. The denominator will stay the same.
whole = x/y;
numerator x%y;
denominator = y;
In case you don't believe me that this works, you can check out http://en.wikipedia.org/wiki/Euclidean_algorithm
I just happen to like the recursive function because it's clean and simple.
Your algorithm is close, but not exactly correct. Also, you should probably create a new function if you want to find the gcd. Just makes it a little cleaner and easier to read. You can also test that function as well.
回答3:
For reference, what you implemented is the original subtractive Euclidean Algorithm to calculate the greatest common divisor of two numbers.
A lot faster version is using the remainder from integer division, e.g. %
instead of -
in your loop:
while (n1 != 0 && n2 != 0){
if(n1 > n2)
n1 = n1 % n2;
else
n2 = n2 % n1;
}
... and then make sure you will use the one which is not zero.
A more streamlined version would be this:
while(n1 != 0) {
int old_n1 = n1;
n1 = n2 % n1;
n2 = old_n1;
}
and then use n1. Matt's answer shows a recursive version of the same algorithm.
回答4:
You should make this class something other than a container for static methods. Here is a skeleton
import java.math.BigInteger;
public class BigRational
{
private BigInteger num;
private BigInteger denom;
public BigRational(BigInteger _num, BigInteger _denom)
{
//put the negative on top
// reduce BigRational using the BigInteger gcd method
}
public BigRational()
{
this(BigInteger.ZERO, BigInteger.ONE);
}
public BigRational add(BigRational that)
{
// return this + that;
}
.
.
.
//etc
}
}
回答5:
I have a similar BigRational
class I use. The GcdFunction
is makes use of BigInteger
's gcd
function:
public class GcdFunction implements BinaryFunction {
@Override
public BigRational apply(final BigRational left, final BigRational right) {
if (!(left.isInteger() && right.isInteger())) {
throw new EvaluationException("GCD can only be applied to integers");
}
return new BigRational(left.getNumerator().gcd((right.getNumerator())));
}
}
BigRational
contains a BigInteger
numerator and denominator. isInteger()
returns true if the simplified ratio's denominator is equal to 1.
来源:https://stackoverflow.com/questions/6618994/simplifying-fractions-in-java