处理一维数组 java

Question

我正在尝试使用对话框中的用户输入创建一个数组。它应该询问用户要输入多少个数字，然后要求用户输入数字。然后代码应该以相反的顺序输出数字。下面是我到目前为止的代码。它不起作用。我在尝试使用用户输入初始化数组时做错了。我是 java 的新手，所以欢迎任何建议。提前致谢。

 public static void main(String[] args)
{
    String input;
    int space;
    double [] numbers;
    double count;
    String numberInput;
    double number;

    input = JOptionPane.showInputDialog
            (null, "How many numbers would you like to enter?");
    space = Integer.parseInt(input);

    numbers = new double[space];

    count = 0;
    while (count < space)
    {
        numberInput = JOptionPane.showInputDialog
                (null, "Enter a number to be sorted: ");
        number = Double.parseDouble(numberInput);

    for (int i = 0; i < numbers.length; i++)
    numbers[i] = number;
        count++;
    }

   double[] numbers2 = swapArray(numbers);
    JOptionPane.showMessageDialog(null, numbers2);
}
public static double[] swapArray(double[] array)
{
    double[] result = new double[array.length];

    for (int i = 0, j = result.length - 1;
            i < array.length; i++, j--)
    {
        result[j] = array[i];
    }
    return result;
}

}

Answer 1

这是我对您的作业的看法。它应该给你一些关于如何处理问题的想法，而不是给你一个复制粘贴的解决方案，也不是最好的（以我自己的能力）structure/logic（但仍然给你提示，让你指出他们的方向):

package so_q33405148;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.logging.Level;
import java.util.logging.Logger;

public class main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        try {
            //TIP: You can use 'Scanner' instead of a 'Reader' here, to avoid having to parse strings into ints (scanner can do it for you)
            BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
            System.out.println("How many numbers would you like to enter?");
            int count = Integer.parseInt(br.readLine());
            int input[] = new int[count];
            for (int i = 0; i < count; i++) {
                System.out.print("Enter a number to be sorted: ");
                //TIP: With some clever math, you can invert the array right as it's still filling up here, by manipulating a new int (so 'i' is unchanged, as it's the for-loop's index) using 'i' and 'input.length'...Or, with a little creativity and insight, you may even achieve the same result manipulating 'i' directly somewhere else...
                input[i] = Integer.parseInt(br.readLine());
            }

            System.out.println("Original array:\n" + Arrays.toString(input));

            //TIP: Better methods to reverse arrays and/or collections exist.
            //Take a look at SO question #3962766 (puritan solution without as much memory-footprint) and also Google about 'Arrays.asList()' and 'Collections.reverse()' (learn about collections-sorting)
            int reversedInput[] = new int[input.length];
            for (int i = 0; i < count; i++) {
                reversedInput[i] = input[count - i - 1];
            }

            System.out.println("Reversed array:\n" + Arrays.toString(reversedInput));
        } catch (IOException ex) {
            Logger.getLogger(main.class.getName()).log(Level.SEVERE, null, ex);
        }
    }
}

处理一维数组 java

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