Java中的数字金字塔

Question

I am a beginner in Java programing and I want to print a pyramid like the one in the image:

I already have the pyramid (without numbers) (x is the numbers of lines the users wants) Any help is good!

for (int i=1; i<(2*x+1); i += 2)
{
    for (int k=0; k < ((x-1) - i / 2); k++)
    {
        System.out.print(" ");
    }
    for (int j=0; j<i; j++)
    {
        System.out.print(x);
    }
    System.out.println("");
}

Answer 1

It might be a bit simpler to write and read if you make a separate method to print each line and cell and then use an if statement to determine what to print. The cell method encapsulates all the logic on what to print.

for (int i = 0; i < size; i++)
    System.out.println(line(i, size));

private String line(int row, int size) {
    StringBuilder line = new StringBuilder();
    for (int col = 0; col < size * 2 + 1; col++) {
        line.append(cell(row, col, size));
    }
    return line;
} 

or if you are using Java 8:

return IntStream.range(0, size * 2 + 1)
    .map(col -> cell(row, col, size)).collect(Collector.joining())))

Finally the method to select the character for each position:

private String cell(int row, int col, int size) {
    int offset = size - col;
    if (offset == row)
        return "0";
    else if (offset > row)
        return " ";
    else if (row == size - 1)
        return "0";
    else
        return row;
}

Answer 2

Without break your existing logic, even though it is hard to read, the code can be(see the comments below):

        for (int i=1; i<(2*x+1); i += 2)
        {
            for (int k=0; k < ((x-1) - i / 2); k++)
            {
                System.out.print(" ");
            }
            for (int j=0; j<i; j++)
            {
                if((j==0)||(j==i-1)){
                    // if it is smallest j or biggest j, print 0
                    System.out.print(0);
                }else if(i == 2*x+1 - 2){
                    // if it is biggest i, print 0
                    System.out.print(0);
                }
                else{
                    // the rest conditions, print row number
                    System.out.print((i-1)/2);
                }
            }
            System.out.println("");
        }

All I did is replace System.out.print(x); with

                    if((j==0)||(j==i-1)){
                        // if it is smallest j or biggest j, print 0
                        System.out.print(0);
                    }else if(i == 2*x+1 - 2){
                        // if it is biggest i, print 0
                        System.out.print(0);
                    }
                    else{
                        // the rest conditions, print row number
                        System.out.print((i-1)/2);
                    }

Answer 3

This is my code for getting the output

for(int i=0;i<x;i++)
{
int space=x-i;
//System.out.println("inside first for");

     for(int j=0;j<(i*2)+1;j++)
     {
         int total_spaces=space;

     if(j==0 || j==(i*2))
     {
      if(j==0)
      {
         while(total_spaces>0)
         {
     System.out.print(" ");
     total_spaces--;
         }
      }

     System.out.print("0");
     if(j==(i*2))
     {
            while(total_spaces>0)
         {
     System.out.print(" ");
     total_spaces--; 
         }
     }
     }


     else
     System.out.print(i);

     }
 System.out.println();
}

Answer 4

Building on the answer by @haifzhan , which is an excellent answer since it gives you a solution with minimal change to the original code, it should be pointed out that the formulas can be simplified if the outer loop is also a simple zero-based increment loop.

The three-way if statement can be replaced with an inline ternary expression ( ?: ), and the code can be enhanced to support pyramids up to a height of 37 by filling the interior with hex digits, as opposed to the current max height of 11.

Compressing the code a bit, you get:

for (int i = 0; i < x; i++) {
    for (int j = 0; j < x - i - 1; j++)
        System.out.print(' ');
    for (int j = 0; j <= i * 2; j++)
        System.out.print(j == 0     ||    // left edge
                         j == i * 2 ||    // right edge
                         i == x - 1 ? '0' // bottom edge
                         : Character.forDigit(i, 36)); // center hex digit
    System.out.println();
}