2 回答

TA贡献1817条经验 获得超6个赞
假设您有一种multiplyBy(Fraction f)方法可以在您的Fraction课堂上正常工作。此外,迭代进行会更好(更具可读性,节省更多时间和空间复杂性)。为了这个例子,我会这样做:
首先定义:
private static boolean oneThirdIsPresent(Fraction[] arr){
for (int i = 0; i < arr.length; i++) {
if(arr[i].numerator == 1 && arr[i].denominator == 3) {
return true;
}
}
return false;
}
private static void recursivelyMultBy2(Fraction[] arr, int index){
if(index == arr.length){
return;
}
arr[index] = arr[index].multiplyBy(new Fraction(2));
recursivelyMultBy2(arr, index+1);
}
为了最终解决:
public static void multBy2IfOneThirdIsPresent(Fraction[] arr){
if(oneThirdIsPresent(arr)){
recursivelyMultBy2(arr, 0);
}else{
return;
}
}

TA贡献1815条经验 获得超13个赞
这是递归乘法部分的一个简单示例:
public static void main(String[] args)
{
Fraction[] fractions = new Fraction[] {new Fraction(1,2), new Fraction(2,3), new Fraction(3,1)};
System.out.println("Fractions:");
for(Fraction f: fractions)
{
System.out.println(f);
}
System.out.println("Multiplying array by 2...");
Fraction.mulAus(fractions, new Fraction(2, 1));
for(Fraction f: fractions)
{
System.out.println(f);
}
}
Modified Fraction Class(乘法代码在底部):
public class Fraction
{
private int numerator; // Zaehler
private int denominator; // Nenner
public Fraction(int num, int denom)
{
if (denom != 0)
{
if (denom < 0)
{
numerator = -num;
denominator = -denom;
}
else
{
numerator = num;
denominator = denom;
}
reduce();
}
else
{
// error: division by zero
//throw new IllegalArgumentException();
}
}
private void reduce()
{
// ...
}
public Fraction()
{
numerator = 0;
denominator = 1;
}
public Fraction(int num)
{
numerator = num;
denominator = 1;
}
public String toString()
{
return numerator + " / " + denominator;
}
public void MultiplyBy(Fraction F)
{
if (F != null)
{
numerator = numerator * F.numerator;
denominator = denominator * F.denominator;
reduce();
}
}
public static void mulAus(Fraction[] arr, Fraction F)
{
if(arr != null && F != null)
{
rekurMul(arr, 0, F);
}
}
private static void rekurMul(Fraction[] arr, int i, Fraction F)
{
arr[i].MultiplyBy(F);
if (i < (arr.length - 1))
{
rekurMul(arr, ++i, F);
}
}
}
输出:
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