Question

i want to generate a pyramid stacks for a given number “n” (“n” – is also a height of the last pyramid). The algorithm should stack (n-1) pyramids. The result should be like this:

height of base pyramid: 4

          2
        2 4 2
          2
        2 4 2
      2 4 8 4 2
          2
        2 4 2
      2 4 8 4 2
   2 4 8 16 8 4 2

I’ve already a code piece to generate last pyramid:

for (i = 1; i <= n; i++) {
    for (j = 1; j <= n - i; j++)
        System.out.printf("%5s", "");
    for (k = 1; k < i; k++)
        System.out.printf("%5d", (int) Math.pow(2, k));
    for (k = i; k >= 1; k--)
        System.out.printf("%5d", (int) Math.pow(2, k));
        System.out.println();
    }

any idea would be helpful.

Answer 1

Add one more for loop in the outer most: –

for (int z = n - 2; z >= 0; z--)

So, your loop structure becomes: –

for (int z = n - 2; z >= 0; z--) {
        for (int i = 1; i <= n - z; i++) {
            for (int j = 1; j <= n - i; j++)
                System.out.printf("%5s", "");
            for (int k = 1; k < i; k++)
                System.out.printf("%5d", (int) Math.pow(2, k));
            for (int k = i; k >= 1; k--)
                System.out.printf("%5d", (int) Math.pow(2, k));
            System.out.println();
        }
    }

z starts with n - 2 because, the first pyramid is of length 2. And note that in the 2nd inner loop, your condition is changed to i <= n - z.

Now, your original set of loops runs 3 times (n - 2) == 2 for n = 4, and loop runs till from z = 2 to z = 0. And it creates a pyramid of height (n - z): – (4 - 2) = 2, (4 - 1) = 3, (4 - 0) = 4 back to back.