Quadratic Read Method

问题

I have to write a read method for a quadratic class where a quadratic is entered in the form ax^2 + bx + c. The description for the class is this:

Add a read method that asks the user for an equation in standard format and set the three instance variables correctly. (So if the user types 3x^2 - x, you set the instance variables to 3, -1, and 0). This will require string processing you have done before. Display the actual equation entered as is and properly labeled as expected output.

I was able to do the ax^2 part by using string manipulation and if else statements. But I am not sure how to do the bx and c parts of the equation because of the sign that could be in front of bx and c. Here is how I did the ax^2 part of the method.

public void read()
{
    Scanner keyboard = new Scanner(System.in);
    System.out.println("Please enter a quadratic equation in standard format.");
    String formula = keyboard.next();
    String a = formula.substring(0, formula.indexOf("x^2"));
    int a2 = Integer.parseInt(a);
    if (a2 == 0)
    {
        System.out.println("a = 0");
    }
    else if (a2 == 1)
    {
        System.out.println("a = 1");
    }
    else
    {
        System.out.println("a = " + a2);
    }
 }

Feel free to write any code as an example. Any help would be greatly appreciated.

回答1:

import java.util.regex.Matcher;
import java.util.regex.Pattern;


public class Mini {

    public static void main(String[] args) {
        int a = 0;
        int b = 0;
        int c = 0;

    String formula = " -x^2 + 6x - 5";
    formula = formula.replaceAll(" ", "");

    if (!formula.startsWith("+") && !formula.startsWith("-"))
        formula = "+" + formula;

        String exp = "^((.*)x\\^2)?((.*)x)?([\\+\\s\\-\\d]*)?$";
        Pattern p = Pattern.compile(exp);
        Matcher m = p.matcher(formula);

        System.out.println("Formula is " + formula);
        System.out.println("Pattern is " + m.pattern());

        while (m.find()) {
            a = getDigit(m.group(2));
            b = getDigit(m.group(4));
            c = getDigit(m.group(5));
        }

        System.out.println("a: " + a + " b: " + b + " c: " + c);

    }

    private static int getDigit(String data) {
        if (data == null) {
            return 0;
        } 
        else 
        {

            if (data.equals("+"))
            {
                return 1;
            }
            else if (data.equals("-"))
            {
                return -1;
            }
            else
            {
                try
                {
                    int num = (int) Float.parseFloat(data);
                    return num;
                }
                catch (NumberFormatException ex)
                {
                    return 0;
                }
            }
        }
    }
}

回答2:

Here's an example of how you could do it using a regular expression. So far this only works properly if the equation is given in the format ax^2 + bx + c. It could be tweaked further to allow for changing the order of the sub-terms, missing terms, etc. To do this I would probably try to come up with regular expressions for each sub-term. Anyway, this should serve to give you the general idea:

import java.util.regex.Pattern;
import java.util.regex.Matcher;

class ParseEquation {
    static Pattern match = Pattern.compile("([\\+\\-]?[0-9]*)x\\^2([\\+\\-]?[0-9]*)x([\\+\\-]?[0-9]*)");

    static String parseEquation(String formula) {
        // remove all whitespace
        formula = formula.replaceAll(" ", "");
        String a = "1";
        String b = "1";
        String c = "0";
        Matcher m = match.matcher(formula);
        if (!m.matches()) return "syntax error";
        a = m.group(1);
        if (a.length() == 0) a = "1";
        if (a.length() == 1 && (a.charAt(0) == '+' || a.charAt(0) == '-')) a += "1";
        b = m.group(2);
        if (b.length() == 0) b = "1";
        if (b.length() == 1 && (b.charAt(0) == '+' || b.charAt(0) == '-')) b += "1";
        c = m.group(3);
        return a + "x^2" + b + "x" + c;
    }

    public static void main(String[] args) {
        System.out.println(parseEquation("2x^2 + 3x - 25"));
        System.out.println(parseEquation("-2x^2 + 3x + 25"));
        System.out.println(parseEquation("+2x^2 + 3x + 25"));
        System.out.println(parseEquation("x^2 + 3x + 25"));
        System.out.println(parseEquation("2x^2 + x + 25"));
    }
}

回答3:

Via regular expressions:

sub quadParse {
    my ($inputStr) = @_;
    my $str = "+".$inputStr;        # as the first needn't have a sign
    $str =~ s/\s+//g;               # normalise
    my $squared = $1 if ($str =~ m/([+-][0-9])*x\^2/);
    my $ex = $1 if ($str =~ m/([+-][0-9]*)x(?!\^)/);
    my $const = $1 if ($str =~ m/([+-][0-9]+)(?!x)/);
    return "${squared}, ${ex}, ${const}";
}

For string parsing, Perl.

Oh, go on then:

public static String coeff(String str, String regex) {
    Pattern patt = Pattern.compile(regex);
    Matcher match = patt.matcher(str);
    // missing coefficient default
    String coeff = "+0"; 
    if(match.find()) 
        coeff = match.group(1);
    // always have sign, handle implicit 1
    return (coeff.length() == 1) ? coeff + "1" 
        : coeff;
}
public static String[] quadParse(String arg) {
    String str = ("+" + arg).replaceAll("\\s", "");
    String quad = coeff(str, "([+-][0-9]*)x\\^2" );
    String ex = coeff(str, "([+-][0-9]*)x(?!\\^)");
    String cnst = coeff(str, "([+-][0-9]+)(?!x)" );
    return new String[] {quad, ex, cnst};
}

Java test in ideone.

These handle the formula in any order, with or without initial sign on the first term, and handle missing terms correctly. The Perl version doesn't fix '+' to '+1' etc, or supply an explicit '0' for missing terms, because I've run out of time.

来源：https://stackoverflow.com/questions/15128758/quadratic-read-method