Question

For a homework assignment, I need help with generalizing the code for printing the following pattern.

   A
  B C  
 D E F 
G H I J

The problem was coding the spaces between the alphabets.
This is what i came up with, but it is just for the first 4 lines of the pattern.
(Sorry i have poor formatting skills >.>)

int r = 65;
char m ;
int count=0;
for(int i = 4;i>0;i--)
{
    for( int j = i;j>0;j--)
        {System.out.print(" ");}
    for(int j = 4-i;j>=0;j--)
        {
             count++;
             m=(char)r;
             if(count == 3||count == 6||count == 8 || count == 11|| count == 13|| count ==15)
             {
                 System.out.print(" ");r--; 

             }
                 else
                     System.out.print(m);r++;
        }
    for(int j = 4-i;j>0;j--)
        {
            count++;m=(char)r;
            if(count == 3||count == 6||count == 8 || count == 11|| count == 13|| count ==15)
            {
                System.out.print(" ");r--;
            }
                    else
                        System.out.print(m);r++;
        }                                                                               
            System.out.println("");
} 

thanks to Gene for the explanation, i did some editing and this is what i came up with.

            int r = 65;
char m ;
for(int i = 4;i>0;i--)
    {
        for( int j = i;j>0;j--)
            {System.out.print(" ");}
        for(int j = 4-i;j>=0;j--)
    {
         m=(char)r;

            System.out.print(m+" ");
            r++;
    }
            System.out.println("");
    }

Answer 1

Writing loops is all about using math (usually simple math) to describe one iteration in terms of loop indicies.

Let N be the number of rows so i=0,1,...N-1 are their indices.

First, your example shows that row i has N-i-1 leading spaces. Let’s check this. For row i=N-1 we get zero and for i=0 we get N-1. In your example the i=0 case is 3. This agrees with the drawing, so we’re looking good.

The second part is that there are i+1 characters per row. All but the last has a following space. The last has a following newline.

Finally, we can get the correct letter by just starting at A and incrementing each time we print a new one.

So now we’re ready to write code:

char ltr = 'A';
for (int i = 0; i < N; i++) {

  // Print the leading spaces.
  int nLeadingSpaces = N - i - 1;
  for (int j = 0; j < nLeadingSpaces; j++) 
    System.out.print(' ');

  // Print the letters for this row.  There are
  // i+1 of them. So print the first i with a following
  // space and the last one with a newline.
  for (int j = 0; j < i; j++) 
    System.out.print(ltr++ + " ");
  System.out.println(ltr++);
}

This is untested but ought to work.