仅打印数组中 3 的倍数

Question

我正在尝试完成一项练习，其中有一些任务，包括以下任务： 仅打印数组中 3 的倍数 我必须使用小程序，但我不知道该怎么做。 我试图在图形部分设置一个条件，但它返回一个不好的 0

public void init() {
  dataList = new int[17];
  int dataList[] = {2,4,6,9,5,4,5,7,12,15,21,32,45,5,6,7,12};

  for (int i = 0; i < dataList.length; i++) {
    //Compute the sum of the elements in the array.
    sum += dataList[i];
    //Compute the product of the elements in the array.
    product *= dataList[i];
    //Compute the frequency of the number 5 in the array
    if (dataList[i] == 5) {
      fiveCounter++;
    }
  }
}

public void paint(Graphics g) {
  g.drawString(("Sum of elements is: " + sum), 25, 25);
  g.drawString(("Product of elements is: " + product), 25, 50);
  g.drawString(("Number 5 is present " + fiveCounter + " times"), 25, 75);
  for (int i = 0; i < dataList.length; i++) {
    if ((dataList[i] % 3) == 0) {
      g.drawString((String.valueOf(dataList[i])), 25, 100);
    }
  }
}

在我尝试基于 3 的倍数的值的数量的计算创建一个新数组的另一次尝试中，程序没有启动，我得到 ArrayIndexOutOfBoundException

public void init() {
  dataList = new int[17];
  multiple3 = new int[mult3Counter];
  int dataList[] = {2,4,6,9,5,4,5,7,12,15,21,32,45,5,6,7,12};

  for (int i = 0; i < dataList.length; i++) {
    //Compute the sum of the elements in the array.
    sum += dataList[i];
    //Compute the product of the elements in the array.
    product *= dataList[i];
    //Compute the frequency of the number 5 in the array
    if (dataList[i] == 5) {
      fiveCounter++;
    }
    if ((dataList[i] % 3) == 0) {
      multiple3[i] = dataList[i];
      mult3Counter++;
    }
  }

public void paint(Graphics g) {
  g.drawString(("Sum of elements is: " + sum), 25, 25);
  g.drawString(("Product of elements is: " + product), 25, 50);
  g.drawString(("Number 5 is present " + fiveCounter + " times"), 25, 75);
  for (int i = 0; i < multiple3.length; i++) {
    g.drawString((String.valueOf(multiple3[i])), 25, 100);
  }
}

我该如何解决这个问题？

Answer 1

您不能对两个数组使用相同的计数器。使用两个不同的计数器。您不能使用mult3Counter 作为数组multiple3 的大小，因为它未初始化！因此，mult3Counter 默认为0。因此，当您要使用任何索引访问multiple3[] 时，它将给出ArayIndexOutOfBoundsException

如果你需要知道3的倍数出现的次数，你必须运行两次循环；

int mult3Counter = 0;
for (int i = 0; i < dataList.length; i++) {
  if ((dataList[i] % 3) == 0) {
    mult3Counter++;
  }
}

int j = 0;
int [] multiple3 = new int[mult3Counter];

for (i = 0; i < dataList.length; i++)
{
  if ((dataList[i] % 3) == 0)
  {
    multiple3[j++] = dataList[i];
  }
}

或者最好的方法是使用List (ArrayList) 将3的倍数相加。

ArrayList<int> multiple3 = new ArrayList<>();

for (int i = 0; i < dataList.length; i++) {
  if ((dataList[i] % 3) == 0) {
    multiple3.add(dataList[i]);
  }
}

如果你需要一个数组，你可以稍后将它转换成一个数组。参考This Question

Answer 2

我自己解决了问题，仅更改了图形部分中的代码。 唯一难看的是我不得不重复 dataList，因为我无法从 init 方法中传输值：

public void paint(Graphics g) {
  int dataList[] = {2,4,6,9,5,4,5,7,12,15,21,32,45,5,6,7,12};
  int yposition = 100;

  g.drawString(("Sum of elements is: " + sum), 25, 25);
  g.drawString(("Product of elements is: " + product), 25, 50);
  g.drawString(("Number 5 is present " + fiveCounter + " times"), 25, 75);
  g.drawString(("The following numbers are multiple of 3"), 25, yposition);

  for (int i = 0; i < dataList.length; i++) {
    if ((dataList[i] % 3) == 0) {
      yposition += 15;
      g.drawString((String.valueOf(dataList[i])), 25, yposition);
    }
  }
}