Print triangle (Half Pyramid) pattern using Java

Question

I have to print a triangle pattern (Half Pyramid) like

1
0 1
1 0 1
0 1 0 1

I tried with this program

class tri{
 public static void main(String arg[]){
    int i,j,a = 1, b =0, c=0;
    for(i=1; i<=4; i++){

        for(j=1; j<=i; j++){
            System.out.print(a+ " ");
            c = a;
            a = b;
            b = c;              
        }
        System.out.println();
    }
 }
}

but this prints pattern as shown in image

please if some one could help me editing that code to bring the pattern

Answer 1

You need to set starting values correctly. Because what you are doing is continuously swapping

Say row two 0 1

last element = 1, (a = 1, b = 0) and on swapping (a = 0, b = 1) for next row first element.

However this is incorrect as it was supposed to start with (a = 1) and not (a = 0) from previous state.

        int i,j,a = 1, b =0, c=0;
        for (i = 1; i <= 4; i++){
            if (i % 2 == 0) {
                a = 0;
                b = 1;          
            } else {
                a = 1;
                b = 0;          
            } 
            for(j=1; j<=i; j++) {
                System.out.print(a+ " ");
                c = a; 
                a = b;
                b = c;                           
            }           
            System.out.println();
        }

You can also switch between 0 and 1 using XOR :

int i, j, a = 1;
for (i = 1; i <= 4; i++){
    a = i % 2;          
    for(j=1; j<=i; j++) {
        System.out.print(a+ " ");
        a = a ^ 1;                         
    }           
    System.out.println();
}

However Shorter solution would be :

String str = "";
for (int i = 1; i <= 4; i++) {
    str = (i % 2) + " " + str;  
    System.out.println(str);    
} 

output :

1 
0 1 
1 0 1 
0 1 0 1 

Answer 2

The shortest form would be

String str = "";
for (int i = 1; i <= 4; i++) {
    str = (i % 2) + " " + str;  
    System.out.println(str);    
} 

This will give output as you desired

1
0 1
1 0 1
0 1 0 1

Answer 3

You can use boolean flag for this to check if you are current starting at 1 or 0;

sample:

boolean flag = true;
    for(int i=1; i<=4; i++){

        for(int j=1; j<=i; j++){
            if(flag)
                System.out.print("1 ");
            else
                System.out.print("0 ");
            flag = !flag;
        }
        if((i % 2) == 0)
            flag = true;
        else
            flag = false;
        System.out.println();
    }

result:

1 
0 1 
1 0 1 
0 1 0 1 

Answer 4

   int x=1,y=1;
   for(int i=1;i<8;i++){
       for(int k=0;k<i;k++){
           y=(k==0) ? x:y;
           System.out.print(y+" ");
           y=(y==1) ? 0:1;
       }
       System.out.println("");
       x=(x==1) ? 0:1;
   }

output---

Answer 5

write my anwser here, seen @sujithvm solution is more short and efficient.

    int sideLength = 4;
    for(int i = 0 ; i < sideLength ; i++)
    {
        for(int j = 0 ; j <= i ; j++)
        {
            System.out.print((i + j + 1) % 2 + " ");
        }
        System.out.println();
    }

Answer 6

Complete Java program for beginners:

public class PrintPattern15

{

public static void main(String args[])
{
    int n = 5;
    PrintPattern15 d = new PrintPattern15();
    d.printPattern(n);
}
public void printPattern(int noOfRows)
{   
    for(int i = 1; i <= noOfRows; i++ )
    {
        printRows(i);
    }
}
public void printRows(int startPt)
{   
    for(int i = startPt ; i >= 1; i--)
    {
        System.out.print(i%2);
    }
        System.out.println();

}

}

Answer 7

This Works for me:

public class DrawPattern {

    public static void main(String[] args) {
        int i, j;
        int num = 7;
        for (i = 0; i <num; i++) {
            for (j = 0; j < i; j++) {
                if (isConditionMatch(num, i, j)) {
                    System.out.print("0");
                } else {
                    System.out.print("1");
                }
            }
            System.out.println();
        }

    }

    private static boolean isConditionMatch(int num, int i, int j) {
        return (i%2 != 0 && j%2 !=0 || (i%2 == 0 && j%2==0));
    }
}

Output:

1
01
101
0101
10101
010101

Answer 8

Code

public class pra1 {
    public static void main(String[] args) {
        int  space, rows=11, k=0;
          //  Scanner scan = new Scanner(System.in);
          //  System.out.print("Enter Number of Rows : ");
          //  rows = scan.nextInt();
        //  rowa=6;

            for(int i=1; i<=rows; i++)
            {
                for(space=1; space<=(rows-i); space++)
                {
                    System.out.print("  ");
                }
                while(k != (2*i-1))
                {  // int p=k;
                    if(k%2==0)
                    {

                    System.out.print("1 ");
                   //  System.out.print("  ");
                    }if(k%2!=0 )
                    {

                    System.out.print("0 ");
                   // System.out.print("  ");
                    }

                    else{
                    System.out.print("");
                    }
                 //   p--;
                    k++;
                }
                k = 0;
                System.out.println();
            }

    }

}

Output

                    1 
                  1 0 1 
                1 0 1 0 1 
              1 0 1 0 1 0 1 
            1 0 1 0 1 0 1 0 1 
          1 0 1 0 1 0 1 0 1 0 1 
        1 0 1 0 1 0 1 0 1 0 1 0 1 
      1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 
    1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 
  1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 
1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 

Answer 9

This is how I solved the problem:

int k = 0;    
for (i = 1; i <= 4; i++){
k = (i%2 == 0)? 1:0;

for(j=1; j<=i; j++) {
    System.out.print(k+ " ");
    k = k==0?1:0;                       
}           
System.out.println();
}