java - 带数字的空菱形

Question

所以我被问到这个问题，我只能解决代码的顶部部分，我被困在底部部分。

Write a Java program called EmptyDiamond.java that contains a method that takes an integer n and prints a empty rhombus on 2n − 1 lines as shown below. Sample output where n = 3:

  1
 2 2
3   3
 2 2
  1

这是我到目前为止的代码:

public static void shape(int n) {
    //TOP PART
    for (int i = 1; i <= (n - 1); i++) {
        System.out.print(" ");
    }
    System.out.println(1);
    for (int i = 2; i <= n; i++) {
        for (int j = 1; j <= (n - i); j++) {
            System.out.print(" ");
        }
        System.out.print(i);
        for (int j = 1; j <= 2 * i - n + 1; j++) {
            System.out.print(" ");
        }
        System.out.println(i);
    }

    //BOTTOM PART (The messed up part)
    for (int i = n + 1; i <= 2 * n - 2; i++) {
        for (int j = 1; j <= n - i; j++) {
            System.out.print(" ");
        }
        System.out.print(i);
        for (int j = 1; j <= n; j++) {
            System.out.print(" ");
        }
        System.out.print(i);
    }
    for (int i = 1; i <= (n - 1); i++) {
        System.out.print(" ");
    }
    System.out.println(1);
}

public static void main(String[] args) {
    shape(4);
}

Answer 1

也许有点晚了，但因为消息的底部部分只是镜像的第一部分，您可以使用 Stack 以相反的顺序打印消息:

public static void main(String[] args) {
    int maxNumber = 3;
    Stack<String> message = new Stack<>();
    // upper part
    for (int row = 0; row < maxNumber; row++) {
        int prefix = maxNumber - (row + 1);
        int spaces = row >= 2 ? row * 2 - 1 : row;

        String line = getLine(row, prefix, spaces);
        System.out.println(line);
        if (row != maxNumber - 1)
            message.add(line);
    }
    // bottom part
    while (!message.isEmpty())
        System.out.println(message.pop());
}

public static String getLine(int row, int prefix, int spaces) {
    StringBuilder line = new StringBuilder("_".repeat(prefix));
    line.append(row + 1);
    if (row != 0) {
        line.append("_".repeat(spaces));
        line.append(row + 1);
    }
    return line.toString();
}

输出:

__1
_2_2
3___3
_2_2
__1

您当然可以使用任何您想要填充堆栈的方法(即生成消息的上部部分)，就像这个问题建议的方法一样。我描述的上半部分包含第一行(包括)到中间行(不包括)。