Code to draw a Big H with all stars

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Background & symptoms

You like letter H ? (?? why I like H??). Anyway, for whatever reasons (to show off, for exam, code test, etc), you need to draw a big H, or two triangles which point to each other. Here is the code for your quick reference:

Output preview

Designing planing

For pattern code test, here are some golden tips

• Normally you’d use two pointers, one for row another for columns , then compare their relationship to determine whether to output or not
• TDD is highly recommended to test boundary case and edge case

Full source code

import java.util.Arrays;
import java.util.Scanner;

public class Pattern {

    public static void main(String[] args) {
        System.out.println("please input a character for display");
        Scanner scanner = new Scanner(System.in);
        String pattern = scanner.nextLine();
        if (pattern == null || pattern.isEmpty()) {
            System.out.println("Invalid input pattern");
            System.exit(1);
        }
        char ptn = pattern.trim().charAt(0);

        System.out.println("please input levels (number smaller than 100)");
        int level = scanner.nextInt();
        if ( level > 100) {
            System.out.println("Invalid input for level");
            System.exit(1);
        }
        char[][] chars = drawPattern(ptn, level);
        print2DArrays(chars);
    }

    private static void print2DArrays(char[][] chars) {
        int row = chars.length;
        for (int i = 0; i < row; i++) {
            int col = chars[i].length;
            for (int j = 0; j < col; j++) {
                System.out.print(chars[i][j]);
            }
            System.out.println();
        }
    }

    private static char[][] drawPattern(char ptn, int level) {
        int len = 2 * level;
        int oppNum = 2 * level - 2;
        char[][] pattern = new char[len - 1][len - 1];

        for (int i = 0; i < len - 1; i++) {
            for (int j = 0; j < len - 1; j++) {
                pattern[i][j] = ' ';
                if (i < level) {
                    if (j <= i || j >= (oppNum - i)) {
                        pattern[i][j] = ptn;
                    }
                } else {
                    if (j <= (oppNum - i) || j >= i) {
                        pattern[i][j] = ptn;
                    }
                }

            }

        }
        return pattern;
    }
}

Code explanation!

Above code is quite self-explanation. However here are some quick note-worthy points:

• You can enter whatever patterns you want to form this big H (not only star)
• Essentially this is using two pointers representing row anc column, then compare their relationship to determine whether to print out or not.

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