Numerical pattern

Question

I want to make a numerical pattern, like the example below. (The input depends on the user input).
Example1:
Input:
6
The required output would be:
1
23
456

Example2:
Input:
7
The required output would be:
1
23
456
7

This is my code:
https://code.sololearn.com/cA0a25a10A12

Answer 1

for(int i = 1; i <= single; i++) {
for(int j = 1; j<=i; j++) {
if(number<=single) //add this
System.out.print(number);
number++;
}
System.out.println();
}

//This works , printing only required numbers .. but for efficient code, should break loop when number >input.. I try to add it later..

edit:
Brianna
perfectly works this.. hope it helps

for(int i = 1; i <= single; i++) {

for(int j = 1; j<=i && number<=single; j++) {
System.out.print(number+" ");
number++;
}
System.out.println();
}

Answer 2

Denise Roßberg hat java einen stream von 1 bis n?

Answer 3

import java.util.Scanner;
class Main{
public static void main(String args[]){
Scanner input= new Scanner(System.in);
int single;
int number = 1;

single= input.nextInt();

for(int i = 1; i <= single; i++) {

for(int j = 1; j<=i; j++) {
System.out.print(number);
number++;
}
if (number==single){
break;
}
System.out.println();
}
}
}

Answer 4

What if you input 7 what is the required output?

(break inner loop when single==number,.. and it works ..)

Answer 5

Example1:When the user inputs: 6.
The output should be:
1
23
456

Example2: When the user inputs: 7
The output should be:
1
23
456
7

Answer 6

Frogged
Yes.

import java.util.stream.IntStream;
for int nums
and LongStream for long nums

IntStream.range(1,n) or to include n you can also use rangeClosed(1,n)

IntStream.rangeClosed(1,6).forEach(i -> {

//do something with i

});