Java 使用HashSet的复数
这所大学的任务是:实现对象的加、减、乘、除方法。从复坐标创建一个尺寸为n的集合(Hashset)。将其转移到一个完成/乘以其元素的方法。 我了解操作方法本身,但我无法理解为什么需要hashset以及如何在代码中实现它。我读了这些文章,但是我仍然不明白如何应用这个类来实现 对不起,我的英语不好Java 使用HashSet的复数,java,hashset,complex-numbers,Java,Hashset,Complex Numbers,这所大学的任务是:实现对象的加、减、乘、除方法。从复坐标创建一个尺寸为n的集合(Hashset)。将其转移到一个完成/乘以其元素的方法。 我了解操作方法本身,但我无法理解为什么需要hashset以及如何在代码中实现它。我读了这些文章,但是我仍然不明白如何应用这个类来实现 对不起,我的英语不好 import java.lang.Math; public class Complex { double dReal, dImaginary; // Constructor method
import java.lang.Math;
public class Complex {
double dReal, dImaginary;
// Constructor methods
public Complex() {}
public Complex( double dReal, double dImaginary ) {
this.dReal = dReal;
this.dImaginary = dImaginary;
}
// Convert complex number to a string
public String toString() {
if (dImaginary >= 0)
return dReal + "+" + dImaginary + "i";
else
return dReal + "-" + -dImaginary + "i";
}
// ================================================================
// Complex number arithmetic
// Compute sum of two complex numbers cA + cB
public Complex Add(Complex cB ) {
Complex sum = new Complex();
sum.dReal = dReal + cB.dReal;
sum.dImaginary = dImaginary + cB.dImaginary;
return (sum);
}
// Compute difference of two complex numbers cA - cB
public Complex Sub( Complex cB ) {
Complex diff = new Complex();
diff.dReal = dReal - cB.dReal;
diff.dImaginary = dImaginary - cB.dImaginary;
return (diff);
}
// Compute product of two complex numbers cA * cB
public Complex Mult( Complex cB ) {
Complex prod = new Complex();
prod.dReal = dReal*cB.dReal - dImaginary*cB.dImaginary;
prod.dImaginary = dImaginary*cB.dReal + dReal*cB.dImaginary;
return (prod);
}
// Compute divisor of two complex numbers cA / cB
public Complex Div( Complex cB ) {
Complex div = new Complex();
double dR, dDen;
if(Math.abs( cB.dReal ) >= Math.abs( cB.dImaginary )) {
dR = cB.dImaginary/cB.dReal;
dDen = cB.dReal + dR*cB.dImaginary;
div.dReal = (dReal + dR*dImaginary)/dDen;
div.dImaginary = (dImaginary - dR*dReal)/dDen;
} else {
dR = cB.dReal/cB.dImaginary;
dDen = cB.dImaginary + dR*cB.dReal;
div.dReal = (dR*dReal + dImaginary)/dDen;
div.dImaginary = (dR*dImaginary - dReal)/dDen;
}
return (div);
}
// ================================================================
// Exercise methods in Complex class
public static void main (String args[]) {
// Setup and print two complex numbers
Complex cA = new Complex( 1.0, 2.0 );
Complex cB = new Complex( 3.0, 4.0 );
System.out.println("cA = " + cA.toString() );
System.out.println("cB = " + cB.toString() );
// Output
Complex cC = cA.Add( cB );
System.out.println("Complex cA + cB = " + cC.toString() );
Complex cD = cA.Sub( cB );
System.out.println("Complex cA - cB = " + cD.toString() );
Complex cE = cA.Mult( cB );
System.out.println("Complex cA * cB = " + cE.toString() );
Complex cF = cA.Div( cB );
System.out.println("Complex cA / cB = " + cF.toString() );
}
}
从复坐标创建一个尺寸为n的集合(Hashset)
它只是存储一组n
复数
将其转移到一个完成/乘以其元素的方法
使用已有的Mult
方法将所有n
复数相乘
Set<Complex> complexNumbers = new HashSet<>();
//add numbers to it
Optional<Complex> result = complexNumbers.stream()
.reduce((c1, c2) -> c1.Mult(c2)); //or .reduce(Complex::Mult)
Set complexNumbers=new HashSet();
//加上数字
可选结果=complexNumbers.stream()
.reduce((c1,c2)->c1.Mult(c2))//or.reduce(复杂::Mult)
注:Java命名惯例是以小写字母开始命名方法(
mult
或multiply
将是合理的名称)关于第一点,如果指令只是“创建一个尺寸为n的集合(Hashset)”,我同意这种解释,但“…来自复杂坐标”的部分会让人困惑。我的猜测是n
(大小)是基于复数的数量,但它仍然只是一个猜测(但可能这只是我对英语的有限理解)。@Pshemo该集合不能用于单独存储实部和虚部(2+2i
-集合中只能有一个2
)。因此我假设它是用来存储实际的复杂对象的。我的猜测是,指令应该听起来更像“…复数坐标的维数n”甚至“…复数的维数n”,而不是“…复数坐标的维数n”@Pshemo Yes。Agreed@Pshemo在俄语中,它听起来像是“复数的n维”,我不知道它在英语中如何听起来更正确。从你的问题来看,在Java约定中,方法应该以小写开头,就像main
,所以将Mult
改为Mult
,将Div
改为Div
。如果你使用完整的单词,比如“代码>乘法< /代码>和