Nested for loop problem.

[Hello]

Hello there, I've been working on this for a while now, and I am still wondering how to solve the problem.
I want to create this:


Here is my code so far,

import java.util.Scanner;

public class Testing {

public static void main (String args[]){
	
	Scanner input = new Scanner(System.in);
	int max;
	
		System.out.print("Max value: ");
		max = input.nextInt();

		printNumbersIncreasing(max);

}	

public static void printNumbersIncreasing(int x){
		
	for (int lineNumber = 1; lineNumber <= x; lineNumber++){
		
		for (int currentField = 1; currentField <= lineNumber; currentField++){
			
			System.out.printf ("%" + lineNumber + "d", currentField);
		}
	
		System.out.println();
	}



}

}

Example output:

Max value: 5
1
 1 2
  1  2  3
   1   2   3   4
    1    2    3    4    5

____

As you can see, my counter starts over each line, how do i get the counter to continue on each line?

Kind Regards Danny.

[tnrh1]

You can print your loops indexes since they reset every loop.
You need another variable that will be raised in 1 after you print him in the nested loop.

for ..
for ..
{
print(counter)
counter++
}

btw: change those long names! I'm using "k" and "i" for my indexes.

[Iron Lion]

btw: change those long names! I'm using "k" and "i" for my indexes.

Nothing wrong with descriptive names for loop counters if it makes the purpose of the loop more clear.

[tnrh1]

Nothing wrong with descriptive names for loop counters if it makes the purpose of the loop more clear.

I didn't said it's wrong.
Using short names helps to read the code faster, save time in writing it and the chance to type it wrong is much lower.
But it's not a demand it's an advice.

for(int LongVariableName=1;LongVariableName<=MaximumNumber;LongVariableName++)
    Sytem.out.println(LongVariableName);

for(int i=0;i<=k;k++)
    System.out.println(i);

Work as you like.

[acmohan]

int k=1;
for(i=1;...;..){
for(j=1;..;..){
if(k<=x)
System.out.printf("%"+i+"d",k);
k++;
}
System.out.println();
}

this will work as u required.this s jst sample. try in this way.jst try implementing

[Hello]

Thanks for helping me out here.

import java.util.Scanner;

public class Boka {


public static void main (String args[]){
	
	Scanner input = new Scanner(System.in);
	int max;
	
		System.out.print("max value: ");
		max = input.nextInt();

		printNumbersIncreasing(max);

} //Main method end	

public static void printNumbersIncreasing(int x){
		
	int counter = 1;
		
		for (int i = 1; i <= x; i++){
		
			for (int j = 1; j <= i; j++){
				if (counter <= x){
					System.out.printf ("%" + i + "d", counter);
					counter++;
			} //if statment end
			
			} // Inner for loop end
	
		System.out.println();

		} //outer for loop end



} //Method end

} //Class end.

Fair enough, it works well, but the only problem now is this:



As you can see on the output pic, i get blank-lines on output. What could be the problem?

Kind Regards Danny.

[tnrh1]

That's cause of the first for loop, he has to stop much before he equals to x.
You can fix it by adding a boolean variable, let say flag and equal it to true.
Then add an "else" to your if statement (counter<=x), and inside the else type flag = false;
Now before the system.out.println(); add an if statement that ask if the the flag is true.(if flag).

Why we did all that?when if (counter <= x) give you false you still printing the new lines and your first for loop keep running.
I'm sure that there is a better solution, I will think about it and edit my post.
The answer is definitely in this line: for (int i = 1; i <= x; i++)

[tnrh1]

After few days I finally got an answer, thanks to 2 guys josAH and pbrockway2.
You asked if counter is <= to x because your loop runned much more times then it should be.
This one is the correct answer:

public static void print (int num)
	{
		int counter = 1;
		for(int i=1;i<=Math.round(Math.sqrt(num*2));i++)
		{
			for(int k=1;k<=i;k++)
			{
				System.out.print(counter+" ");
				counter++;
			}
			System.out.println();
		}
			
	}

This loop will run as many times as it needs.