绘制数字菱形 [英] Drawing numeric diamond

问题描述

我需要绘制一个数字菱形，例如高度为 9 :

I need to draw a numeric diamond, for example with a height of 9:

    1
   222
  33333
 4444444
555555555
 4444444
  33333
   222
    1

我编写了代码，并设法获得了相同的钻石，但带有星星.我想要这些数字.我怎样才能做到这一点?到目前为止，这是我所做的:

I wrote the code and I managed to get the same diamond, but with stars. I want it with these numbers. How can I do that? Here is what I have done so far:

for (int i = 1; i < 10; i += 2) {
    for (int j = 0; j < 9 - i / 2; j++)
        System.out.print(" ");
    for (int j = 0; j < i; j++)
        System.out.print("a");
    System.out.print("\n");
}

for (int i = 7; i > 0; i -= 2) {
    for (int j = 0; j < 9 - i / 2; j++)
        System.out.print(" ");
    for (int j = 0; j < i; j++)
        System.out.print("b");
    System.out.print("\n");
}

推荐答案

关于您的代码:

• System.out.print("\ n"); 应该替换为 System.out.println().
• 您应该使用动态高度，而不要硬编码9.
• 它会打印正确的图案，只是打印出的图案是错误的:您应该打印循环的索引，然后打印循环代码，而不是打印"a" 和"b" 看看你能从那里得到什么.这是@ Tsung-Ting Kuo解决方案.

• System.out.print("\n"); should be replaced by System.out.println().
• You should a dynamic height instead of hard-coding 9.
• It prints the correct pattern, only what is printed is wrong: instead of printing "a" and "b", you should print the index of the loop and see what you can get from there. This is @Tsung-Ting Kuo solution.

在我看来，您可以用更少的循环和更容易理解的方式来做到这一点.考虑以下算法:

You can do it with fewer loops and a more understandable in my view. Consider the following algorithm:

• 对于模式的每一行(因此该行从0变为 height )
• 对于模式的每一列(因此该列从0变为 height )
• 在图表的右上角，左上角，右下角或左下角时，我们需要打印一个空格.
  • 左上:这是当列小于 height/2-row-1
  时
  • 左下:这是当列小于 row-height/2 时
    • 将这两个表达式组合为一个表达式，这是当列小于 height/2-min 时，其中 min = Math.min(row + 1，height-row).
    • For each row of the pattern (so the row goes from 0 to height excluded)
    • For each column of the pattern (so the column goes from 0 to height excluded)
    • We need to print a space when we are in the upper-right, upper-left, lower-right or lower-left of the diagram.
      • Upper-left: This is when the column is less than height/2-row-1
      • Lower-left: This is when the column is less than row-height/2
        • Factoring these two expressions in a single one, this is when the column is less than height/2 - min where min = Math.min(row+1, height-row).
        • 将这两个表达式组合为一个表达式，这是当列大于 height/2 + min 时，其中 min = Math.min(row + 1，height-row).

        变成代码:

        public static void main(String[] args) {
            int height = 9;
            for (int row = 0; row < height; row++) {
                for (int column = 0; column < height; column++) {
                    int min = Math.min(row+1, height-row);
                    if (column <= height / 2 - min || column >= height / 2 + min) {
                        System.out.print(" ");
                    } else {
                        System.out.print(min);
                    }
                }
                System.out.println();
            }
        }
        

        示例输出:

            1    
           222   
          33333  
         4444444 
        555555555
         4444444 
          33333  
           222   
            1    
        

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