Java程序从二次方程中提取系数

时间:2015-04-18 16:51:58

标签: java regex string math

问题:Java程序将系数从二次方程中分离,例如 如果输入字符串是:

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

所以我需要从给定的输入字符串中分割系数并将输出作为

输出
a = 4 b = -4 c = -42

我试过了:

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]);

但我的输出为23x2和4x和4。 所需的实际输出为23 ,- 4 , 4

5 个答案:

答案 0 :(得分:2)

使用Regex,以下模式将起作用:

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

这将匹配二次方并提取参数,让我们弄清楚它是如何工作的:

  • (...)这是捕获群组
  • [+-]?\d+这匹配了多个数字,前面可选+-
  • [Xx]这匹配&#34; X&#34;或&#34; x&#34;
  • \s*匹配零个或多个空格

所以

  • ([+-]?\d+)匹配&#34; a&#34;参数
  • [Xx]2匹配&#34; X2&#34;或&#34; x2&#34;
  • \s*匹配可选空格
  • ([+-]?\d+)匹配&#34; b&#34;参数
  • [Xx]匹配&#34; X&#34;或&#34; x&#34;
  • \s*匹配可选空格
  • ([+-]?\d+)匹配&#34; c&#34;参数
  • \s*=\s*0匹配&#34; = 0&#34;有一些可选空格

让我们将其包装在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();
    }
}

请注意\\,这是Java所必需的。

快速测试:

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

输出:

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

答案 1 :(得分:0)

如果您只使用样方:

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);

我可能搞砸了子串,但你得到了一般的想法。

答案 2 :(得分:0)

您可以按如下方式使用正则表达式:

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);
}

输出:

a=4 b=-4 c=-42 

答案 3 :(得分:0)

第一个注意事项:如果在regexp中使用符号作为分隔符,则会在slpitted元素中丢失它。我建议你使用以下正则表达式:

"x2|x|="

然后,你只能获得数字。 完整的代码片段是:

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]));
        }
    }
}

但在这种情况下,输出中会有'+'符号。为避免这种情况,您可以使用Integer.parseInt(parts[i])代替parts[i]

答案 4 :(得分:0)

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