java - Java 数组多项式加减法

Question

我有下面的代码，一切似乎都工作正常，除了尝试从另一个多项式中减去一个多项式时的 sub() 函数

p4 = p1.sub(p2);

我看到输出:x+3x^5-5x^8，当我期望看到:x-2x^3+7x^5-2x^7-5x^ 8..我无法弄清楚我的代码有什么问题，非常感谢任何帮助。提前致谢。

public class Polynomial {

public static final int MAX_NUMBER_OF_COEFFICIENTS = 30;
private int[] coefficients = new int[MAX_NUMBER_OF_COEFFICIENTS];

public Polynomial() {
}

public Polynomial(int coefficient, int exponent) {
    coefficients[exponent] += coefficient;
}

public Polynomial(int[] newcoefficients) {
    for (int i = 0 ; i < MAX_NUMBER_OF_COEFFICIENTS; i++) {
        coefficients[i] += newcoefficients[i];
    }
}

public void insert(int coefficient, int exponent) {
    coefficients[exponent] += coefficient;
}

public Polynomial add(Polynomial otherPolynomial) {
    for (int i = 0 ; i < MAX_NUMBER_OF_COEFFICIENTS; i++) {
        if (otherPolynomial.coefficients[i] != 0)
            this.coefficients[i] += otherPolynomial.coefficients[i];
    }
    return new Polynomial(this.coefficients);
}

public Polynomial sub(Polynomial otherPolynomial) {
    for (int i = 0 ; i < MAX_NUMBER_OF_COEFFICIENTS; i++) {
        if (otherPolynomial.coefficients[i] != 0)
            this.coefficients[i] -= otherPolynomial.coefficients[i];
    }
    return new Polynomial(this.coefficients);
}

public void sub(int coefficient, int exponent) {
    this.coefficients[exponent] -= coefficient;
}

public void remove(int exponent) {
    coefficients[exponent] = 0;
}

public void printout() {
    boolean firstPrint = false;
    if (coefficients[0] != 0) {
        System.out.print(coefficients[0]);
        firstPrint = true;
    }
    for (int i = 1 ; i < MAX_NUMBER_OF_COEFFICIENTS; i ++) {
        if (coefficients[i] != 0) {
            if (firstPrint == false) {
                System.out.print(coefficients[i] + "x^" + i);
                firstPrint = true;
            }
            else {
                System.out.print(" + " + coefficients[i] + "x^" + i);
            }
        }
    }
    System.out.println();
}

public static void main(String[] args) {

    Polynomial p1 = new Polynomial(1,1);
    Polynomial p2 = new Polynomial(2,3);
    Polynomial p3 = new Polynomial();
    Polynomial p4 = new Polynomial();
    Polynomial p5 = new Polynomial(6,3);

    try {
        p1.insert(3,5);
        p1.sub(5,8);
        p1.printout();
        p2.sub(4,5);
        p2.add(new Polynomial(2,7));
        p2.printout();
        p3 = p1.add(p2);
        p3.printout();
        p4 = p1.sub(p2);
        p4.printout();
        p5.insert(2,5);
        p5.remove(3);
        p5.printout();
        //p1.add(new Polynomial(2,-1));
    }
    catch (ArrayIndexOutOfBoundsException e) {
            System.out.println("Exponent cannot be negative. " + e);
    }
}
}

Answer 1

该功能正在按预期工作。您不知道(也许您知道)的是，当您调用 p3 = p1.add(p2); 时，您将更改 p1 ，然后返回一个新的多项式p1 的系数。因此，当您调用 p4 = p1.sub(p2) 时，您将获得与 p3 = p1.add(p2) 之前相同的 p1 >。所以你所拥有的更像是这样的:

p3 = p1* = p1 + p2

p4 = p1** = p1* - p2

其中p1*是p1 + p2的结果。所以你现在拥有的是:

p4 = p1* - p2 = (p1 + p2) - p2 = p1

再次强调，如果您想返回新的多项式，请不要更改现有的多项式。

更改两个函数add(Polynomial ..)和sub(Polynomial ..):

public Polynomial add(Polynomial otherPolynomial) {
    int coeff[] = new int[MAX_NUMBER_OF_COEFFICIENTS];
    System.arraycopy(this.coefficients, 0, coeff, 0, MAX_NUMBER_OF_COEFFICIENTS);
    for (int i = 0 ; i < MAX_NUMBER_OF_COEFFICIENTS; i++) {
        // why is this anyway?
        //if (otherPolynomial.coefficients[i] != 0)
        coeff[i] += otherPolynomial.coefficients[i];
    }
    return new Polynomial(coeff);
}

public Polynomial sub(Polynomial otherPolynomial) {
    int coeff[] = new int[MAX_NUMBER_OF_COEFFICIENTS];
    System.arraycopy(this.coefficients, 0, coeff, 0, MAX_NUMBER_OF_COEFFICIENTS);
    for (int i = 0 ; i < MAX_NUMBER_OF_COEFFICIENTS; i++) {
        // why is this anyway?
        //if (otherPolynomial.coefficients[i] != 0)
        coeff[i] -= otherPolynomial.coefficients[i];
    }
    return new Polynomial(coeff);
}

这样，当您分配(或调用复制构造函数)p3 和 p4 时，您就不会更改 p1。

<小时/>

编辑:根据您的代码，不可能有一个函数一旦添加到原始调用者并且一旦不更改它，这正是您的程序想要的。当您调用 p2.add(new Polynomial(2,7)); 时，您想要更改 p2，同时您想要 p1.add(p2 ) 不要改变 p1 ，这是不可能在一个函数中实现的。

如果您想获得问题中建议的输出，请更改

p2.add(new Polynomial(2,7));

到

p2 = p2.add(new Polynomial(2,7));

<小时/>

EDIT2:如果我想这样做，我真的会以完全不同的方式来做。我将拥有所有具有 void 返回类型的 add/sub 函数，并且有两个返回新 Polynomial< 的静态 add/sub 函数 并接受两个多项式。这些静态方法不能在实例上调用(如p1)，而只能在类上调用。

因此，如果我设计的话，这将是我的代码:

public static Polynomial add(Polynomial p1, Polynomial p2){
    Polynomial newPolynomial = new Polynomial();
    for(int i=0;i< MAX_NUMBER_OF_COEFFICIENTS;i++)
    newPolynomial.coefficients[i] = p1.coefficients[i]+p2.coefficients[i];
    return newPolynomial;
}

public static Polynomial sub(Polynomial p1, Polynomial p2){
    Polynomial newPolynomial = new Polynomial();
    for(int i=0;i< MAX_NUMBER_OF_COEFFICIENTS;i++)
        newPolynomial.coefficients[i] = p1.coefficients[i]-p2.coefficients[i];
    return newPolynomial;
}

public void add(Polynomial otherPolynomial) {
    for (int i = 0 ; i < MAX_NUMBER_OF_COEFFICIENTS; i++) {
        coefficients[i] += otherPolynomial.coefficients[i];
    }
}

public void sub(Polynomial otherPolynomial) {
    for (int i = 0 ; i < MAX_NUMBER_OF_COEFFICIENTS; i++) {
        coefficients[i] -= otherPolynomial.coefficients[i];
    }
}

当你这样做时，你不能说p3.add(p1, p2)。这会将您的主要代码更改为:

    p1.insert(3,5);
    p1.sub(5,8);
    p1.printout();
    p2.sub(4,5);
    p2.add(new Polynomial(2,7));
    p2.printout();
    p3 = Polynomial.add(p1, p2);
    p3.printout();
    p4 = Polynomial.sub(p1, p2);
    p4.printout();
    p5.insert(2,5);
    p5.remove(3);
    p5.printout();

如您所见，对于 p3 和 p4，您将调用 Polynomial.add 和 Polynomial.sub 。您可以通过理解 Math.* 方法来最好地理解这一点。所有 Math 库都是静态，因此您无法创建它的实例，但您可以说 Math.abs...这里是相同的概念，你可以调用Polynomial.add等等