java - 用三角力量打造三角力量？

Question

我已经编写了一段代码，使用 for 循环创建完美的三角力。我如何使用更多的 for 循环来从这些三角力中创建三角力？

注意:“ROptionPane”与“JOptionPane”相同

public static void main(String[] args) {


    String rowsWantedAsSt = ROptionPane.showInputDialog(null, "How big would you like the Diamond to be?",
            "Diamond", ROptionPane.QUESTION_MESSAGE);

    int rowsWanted = Integer.parseInt(rowsWantedAsSt);


    //stars on the top row
    int starsWanted = 1;
    //spaces on the top row
    int spacesWanted = (rowsWanted * 2) + 3;





    for (int rows = 0; rows < rowsWanted; rows++) {
            for (int spaces = 0; spaces < spacesWanted; spaces++) {
                System.out.print(" ");
            }
            for (int stars = 0; stars < starsWanted; stars++) {
                System.out.print("*");
            }
            System.out.println();
            starsWanted += 2;
            spacesWanted--;
    }
            starsWanted = 1;

            spacesWanted = rowsWanted + 1;

        for (int rows = 0; rows < rowsWanted; rows++) {
            for (int spaces = 0; spaces < spacesWanted + 2; spaces++) {
                System.out.print(" ");
            }
            for (int stars = 0; stars < starsWanted; stars++) {
                System.out.print("*");
            }

            for (double spaces = 0; spaces < (spacesWanted * 2) - 3; spaces++) {
                System.out.print(" ");
            }
            for (int stars = 0; stars < starsWanted; stars++) {
                System.out.print("*");
            }
            System.out.println();
            starsWanted += 2;
            spacesWanted--;

        }      
    }
}

Answer 1

Triforce 与创建Sierpinski triangle 的算法第一次迭代的结果相同。 。由 Triforce 组成的 Triforce 看起来像是该算法的第二次迭代。

Rosettacode.org 有两个不同的 Java implementations迭代生成 ASCII Sierpinski 三角形的函数；您可能想从这里开始。