Java程序从二次方程中提取系数

Question

Problem: Java program to split the coefficients from a quadratic equation eg if input string is:

String str1;
str1 = "4x2-4x-42=0"

So I need to split the coefficients from the given input string and to get output as

a = 4 b = -4 c = -42

I tried this:

String equation = "ax2+bx-c=0";
String[] parts = equation.split("\\+|-|=");
for (int i = 0; i < parts.length - 2; i++) {
    String part = parts[i].toLowerCase();
    System.out.println(part.substring(0, part.indexOf("x")));
}
System.out.println(parts[2]);

But I got the output as 23x2 and 4x and 4. Actual output needed is 23 ,- 4 , 4 .

Answer 1

Use Regex, the following pattern will work:

([+-]?\d+)[Xx]2\s*([+-]?\d+)[Xx]\s*([+-]?\d+)\s*=\s*0

This will match the quadratic and extract the parameters, lets work out how it works:

• (...) this is capturing group
• [+-]?\\d+ this matches a number of digits, preceded optionally by a + or -
• [Xx] this matches "X" or "x"
• \\s* this matches zero or more spaces

So

• ([+-]?\\d+) matches the "a" argument
• [Xx]2 matches "X2" or "x2"
• \\s* matches optional whitespace
• ([+-]?\\d+) matches the "b" argument
• [Xx] matches "X" or "x"
• \\s* matches optional whitespace
• ([+-]?\\d+) matches the "c" argument
• \\s*=\\s*0 matches "=0" with some optional spaces

Lets wrap this in a class :

private static final class QuadraticEq {
    private static final Pattern EQN = Pattern.compile("([+-]?\\d+)[Xx]2\\s*([+-]?\\d+)[Xx]\\s*([+-]?\\d+)\\s*=\\s*0");
    private final int a;
    private final int b;
    private final int c;

    private QuadraticEq(int a, int b, int c) {
        this.a = a;
        this.b = b;
        this.c = c;
    }

    public static QuadraticEq parseString(final String eq) {
        final Matcher matcher = EQN.matcher(eq);
        if (!matcher.matches()) {
            throw new IllegalArgumentException("Not a valid pattern " + eq);
        }
        final int a = Integer.parseInt(matcher.group(1));
        final int b = Integer.parseInt(matcher.group(2));
        final int c = Integer.parseInt(matcher.group(3));
        return new QuadraticEq(a, b, c);
    }

    @Override
    public String toString() {
        final StringBuilder sb = new StringBuilder("QuadraticEq{");
        sb.append("a=").append(a);
        sb.append(", b=").append(b);
        sb.append(", c=").append(c);
        sb.append('}');
        return sb.toString();
    }
}

Note the \\\\ , this is required by Java.

A quick test:

System.out.println(QuadraticEq.parseString("4x2-4x-42=0"));

Output:

QuadraticEq{a=4, b=-4, c=-42}

Answer 2

If you are only using quadratics:

int xsqrd = equation.indexOf("x2");
int x = equation.indexOf("x", xsqrd);
int equ = equation.indexOf("=");
String a = equation.subString(0,xsqrd);
String b = equation.subString(xsqrd+1,x);
String c = equation.subString(x,equ);

I may have messed up the substrings but you get the general idea.

Answer 3

You can use a regex as follows:

final String regex = "([+-]?\\d+)x2([+-]\\d+)x([+-]\\d+)=0";
Pattern pattern = Pattern.compile(regex);

final String equation = "4x2-4x-42=0";
Matcher matcher = pattern.matcher(equation);

if (matcher.matches()) {
    int a = Integer.parseInt(matcher.group(1));
    int b = Integer.parseInt(matcher.group(2));
    int c = Integer.parseInt(matcher.group(3));
    System.out.println("a=" + a + " b=" + b + " c=" + c);
}

Output:

a=4 b=-4 c=-42 

Answer 4

The first note: if you use symbol as delimiter in regexp, you will lose it in slpitted elements. I suggest you use the folllowing regexp:

"x2|x|="

Then, you can got only numbers. The full fragment of code is:

public class Main {
    private static final char[] varNames = {'a', 'b', 'c'};

    public static void main(String[] args) {
        String equation = "4x2-4x-42=0";
        String[] parts = equation.split("x2|x|=");
// you will get 4 elements, but the last element is always 0
        for(int i=0; i<parts.length - 1; i++){
            System.out.println(varNames[i] + " = " + Integer.parseInt(parts[i]));
        }
    }
}

But in this case you will have the '+' symbols in output. To avoid it, you may use Integer.parseInt(parts[i]) instead of parts[i] .

Answer 5

for ( int i = 0 ; i < str.length ; ++i ){
  if(asciiOf(str.charAt(i)) == asciiOfIntegerValue ){
    addCharInArrayList(str.charAt(i));
  }else{
      addInFuncList();
      addCharInArrayList("_");
  }


  // join numbers which are not separated by _ and apply BODMAS rule and solve it  
  // fyi : apologies - very inefficient pseudocode, wrote in a haste