# Java Program to Print Double Sided Stair Case Number Pattern

## Program to Print Double Sided Stair Case Number Pattern

In the previous article, we have discussed Java Program to Print Stair Case Number Pattern

In this article we are going to see how to print double sided staircase number program.

Example-1

When row value=8

0 1
0 1
0 1 2 3
0 1 2 3
0 1 2 3 4 5
0 1 2 3 4 5
0 1 2 3 4 5 6 7
0 1 2 3 4 5 6 7
Example-2:

When row value= 6

0 1
0 1
0 1 2 3
0 1 2 3
0 1 2 3 4 5
0 1 2 3 4 5

Now, let’s see the actual program to print it.

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Approach:

• Enter total row and store it in an integer variable row.
• Take first outer for loop to keep track of number of rows.
• Take first inner for loop to print spaces .
• Take second inner for loop for printing numbers.
• Then go on printing the number according to loop.

### Java Code to Print Double Sided Stair Case Number Pattern

import java.util.*;
public class Main
{
public static void main(String args[])
{
// taking variable for loop iteration and row .
int row ,c,r,k;
//creating object
Scanner s = new Scanner(System.in);
// entering the number of row
System.out.print("Enter rows : ");
row = s.nextInt();
for (r = 1; r <= row; r++)
{
if(r % 2 != 0)
k = r + 1 ;
else
k = r;
//  loop for printing spaces
for (c = row; c > k; c--)
System.out.print(" ");
//  loop for printing numbers
for (c = 0; c < k; c++)
System.out.print(c+" ");
System.out.println();
}
}
}
Output:

Enter rows : 6

0 1
0 1
0 1 2 3
0 1 2 3
0 1 2 3 4 5
0 1 2 3 4 5

### C Code to Print Double Sided Stair Case Number Pattern

#include <stdio.h>
int main()
{
int row,r,c,k ;
printf("Enter rows: ");
scanf("%d", &row);
for (r = 1; r <= row; r++)
{
if(r % 2 != 0)
k = r + 1 ;
else
k = r;
for (c = row; c > k; c--)
printf(" ");
for (c = 0; c < k; c++)
printf("%d ",c);
printf("\n");
}
return 0;
}
Output:

Enter rows : 6

0 1
0 1
0 1 2 3
0 1 2 3
0 1 2 3 4 5
0 1 2 3 4 5

### C++ Code to Print Double Sided Stair Case Number Pattern

#include <iostream>
using namespace std;
int main()
{
int row,r,c,k ;
cout << "Enter rows: ";
cin>> row;
for (r = 1; r <= row; r++)
{
if(r % 2 != 0)
k = r + 1 ;
else
k = r;
for (c = row; c > k; c--)
cout <<" ";
for (c = 0; c < k; c++)
cout << c << " ";
cout <<"\n";
}
return 0;
}
Output:

Enter rows :  6

0 1
0 1
0 1 2 3
0 1 2 3
0 1 2 3 4 5
0 1 2 3 4 5

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