java - 我怎样才能让我的程序返回-1/4而不是1/-4？

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\帮帮我！我的作业要在今晚12点前完成，我似乎无法让我的程序输出负数为-1/4而不是1 / -4！这让我很沮丧，因为事实是，我不能只在数字前加一个“-”（正如您在下面的程序中看到的那样，我已将其注释掉了）。
我试图将分数设为-1/4（仅以该分数为例）是在“ public RationalNumber（int ...）{行下：您看到我写有* -1的部分了吗？在导致我得到1 / -4的那一步出错了？这让我感到沮丧...在此先感谢。

private int n;
private int d;
public RationalNumber(int numerator, int denominator) {
    if (denominator == 0) {
        throw new IllegalArgumentException();
    }
    if (denominator < 0) {
        denominator = denominator * -1;
        numerator = numerator * -1;
      }
    n = numerator;
    d = denominator;
    simplified();
}

public RationalNumber add(RationalNumber rn) {
    int comDenom = d * rn.getDenominator();
    int num1 = n * rn.d;
    int num2 = rn.n * d;
    d = comDenom;
    return new RationalNumber(num1 + num2, d);
}

public RationalNumber subtract(RationalNumber rn) {
    int num1 = n * rn.d;
    int num2 = rn.n * d;
    return new RationalNumber(num1 - num2, d);
}

public RationalNumber multiply(RationalNumber rn) {
    int numer = n * rn.n;
    int denom = d * rn.d;
    return new RationalNumber(numer, denom);
}

public RationalNumber divide(RationalNumber rn) {
    int numer = n * rn.d;
    int denom = d * rn.n;
    return new RationalNumber(numer, denom);
}

public RationalNumber() {
    n = 0;
    d = 1;
}

public boolean equals(RationalNumber o) {
    return (n == o.getNumerator() && d == o.getDenominator());
}

public int getDenominator() {
    return d;
}

public int getNumerator() {
    return n;
}

public String toString() { //Prints everything here.
    if (d == 1) {
        return "" + n;
    }
    if (n == 0) {
        return "0";
    }
    /*if ((n < 0) || (d < 0)) {
        return "-" + Math.abs(n) + "/" + Math.abs(d);
    }*/
    return n + "/" + d;
}

private void simplified() {
     if (n != 0) {
         int cDenom = gcd(n, d);
         n = n / cDenom;
         d = d / cDenom;
     }
}

private int gcd(int gcdNum1, int gcdNum2) {
    while (gcdNum1 != 0 && gcdNum2 != 0) {
        int tempNum = gcdNum2;
        gcdNum2 = gcdNum1 % gcdNum2;
        gcdNum1 = tempNum;
    }
    return gcdNum1 + gcdNum2;
}

最佳答案

问题是gcd()方法可能返回否定结果。调用simplified()方法时，将n和c除以可能为负的数字，然后更改其符号。要解决此问题，可以更改gcd()方法以始终返回肯定结果。

更改此行：

return gcdNum1 + gcdNum2;

至

if (gcdNum1 + gcdNum2 < 0) {
    return -(gcdNum1 + gcdNum2);
} else {
    return gcdNum1 + gcdNum2;
}

然后检查是否得到正确的结果。