number pattern programs in java

Question

How to print the triangle below:

2  3  5  8  3  8
   4  6  9  4  9
      7  1  5  1
         2  6  2
            7  3
               4

First you need to start with number 2 and add one to the next one vertically

My code:

        int d = 2, n = 6;
        for (int line=1; line <= n; line++ ) {
            for (int j = 2; j <= line; j++) {
                System.out.print("  ");
            }
            for (int k = line; k <= n; k++) {
                System.out.print(d + " ");
                    d = d + k;
                    if (d > 9) {
                        d = d - 9;
                    }
            }
            System.out.println();
        }

Result:

2 3 5 8 3 8 
  5 7 1 5 1 
    7 1 5 1 
      7 2 7 
        4 9 
          6 

Answer 1

The pattern is that the value of d has to be calculated initially on every new line based on the value of d in the first instance of the previous line . That is the part that's missed here. You can do that by having a temp variable store the initial value of d on every line and print based on that. I have used a variable tempD here, which can help print the pattern that you require.

    int d = 2, n = 6;
    int tempD = d - 1;
    for (int line = 1; line <= n; line++) {
        tempD = tempD + line;
        if (tempD > 9) {
            tempD = tempD - 9;
        }
        d = tempD;
        for (int j = 2; j <= line; j++) {
            System.out.print("  ");
        }
        for (int k = line; k <= n; k++) {
            System.out.print(d + " ");
            d = d + k;
            if (d > 9) {
                d = d - 9;
            }
        }
        System.out.println();
    }