Java Diamond with Numbers

Determine the number of lines to output (2*n + 1). then for each line:

• determine how many leading spaces (m) and the highest digit t
• print m spaces
• print the integers from 0 through t
• print the digits from t-1 down to zero
• print a newline

Here's code that works for n=0..15

public static void printTriangle(int n ) {
  int lines = (2 * n) + 1;
  for (int i = 0; i < lines; i++) {
     int nsp;
     int top;
     if ( i <= n ) {
        nsp = n - i;
        top = i;
     } else {
        nsp = i - n;
        top = n - nsp;
     }
     // print out nsp spaces
     for (int j = 0; j < nsp; j++) {
        System.out.print( ' ' );
     }

     // print integers 0 through top
     for (int j = 0; j <= top; j++) {
        System.out.printf( "%x", j );
     }
     // ptint numbers from top-1 to zero
     for (int j = top - 1; j >= 0; j--) {
        System.out.printf( "%x", j );
     }
     System.out.println( "" );
  }
}

I would approach this a two step problem. Build the pattern, then print the pattern. When you try to build it as you print it, you are forced to think left-to-right and then top-to-bottom. Instead of being able to formulate an appropriate mathematical model.

Abhinav Pandey's answer is a good solution for the n <= 9 limit, but there are other models beyond that. You know that the resultant matrix will be 2n-1 by 2n-1, so allocate an array of that size and then fill it in by starting in the center and working your way out.

Printing it out is now a very direct nested loop.

The diamond is simmetric, so in pseudocode:

Print_diamond = do {print top; print center; print reverse(top)}

The center is easy to generate:

0.upto(n) + (n - 1).downto(0)

Then reduce it step by step by replacing the center 3 items with the center item minus one until you arrive to a single 0:

Example: 01210 
          010
121 becomes 010
010 becomes 0
Stop

The number of spaces is equal to the max line length - current line length, half before, half after the digits.

Separate this steps into functions and you will have clear code.

When coding create a plan, divide it into steps and then write the actual code. Writing many nested for loops and conditions without an overall view rarely works out for more complex problems (even if it does it is fragile to the addition of functionality and changes).

Before going into any coding I would give you a quick hack for the diamond problem . Instead of making so many for looping and columns and keeping track of variables the diamond problem is a variant of a maths problem. HERE

1*1 = 1
11*11 = 121
111*111 = 12321
1111*1111 = 1234321 

and so on. SO if you use this formula you just need one for loop for spaces and one for increasing your num*10 +1 where num is 1 and then squaring it and you will get the diamond pattern. If you want I can code it for you but since you asked for an easier method this is the best you can get.

Solution suggested by FredK is something I would implement, as it breaks down the problem to more smaller problems. However, this only prints the upper part of the diamond, so I would suggest doing the following:

• don't immidiately print characters out, store them in a String and when a line is ready, store it in an array, or a List, called, for example, lines

• when you are done with upper half of the diamond, print all the lines, each one in a new line

• reverse the lines and remove the first element (to avoid duplicating the middle line)

• print the reversed lines for the last time

This method is immediately smelly:

public static void printNumberDiamond(int diaCenter)
{   
        printNumberDiamondTop(diaCenter);
        printNumberDiamondBottom(diaCenter);
}

Which of the called methods will print the center line? It's impossible to tell without looking at the implementation, and you don't want readers to have to ask such questions. You want to make the logic as obvious as possible.

A simple step that will solve this issue is to add a third method to print the center line, and call it between these two methods.

Then, as a next step, realize that since the bottom is a mirror of the top, the logic to print them should have some common element. Looking at this code, or looks like there is no common element, and it will be a good idea to reorganize the code to make that happen.