How do I do this? 1 1. 2. 1 1. 2. 3. 2. 1 1. 2. 3. 4. 3. 2. 1

Answer 1

import java.util.Arrays;
public class Program {
public static void main(String[] args) {
int r = 4;

char[] chars = new char[r];
Arrays.fill(chars, ' ');
String spaces = new String(chars);

for (int i=0; i < r; i++){
// print space
System.out.print(spaces.substring(i));

// print left number
for(int j=1; j<=i+1; j++){
System.out.printf("%2d",j);
}

// print right number
for(int j=i; j > 0; j--){
System.out.printf("%2d",j);
}

// print new line
System.out.println();
}
}
}

Answer 2

how can I do this?

Answer 3

If you understand this C code, you can do it with Java. This triangle called Pascal triangle.

#include <stdio.h>

long fact(int);

int main() {

int line, i, j;

printf("Enter the no. of lines: ");
scanf("%d", &line);

for (i = 0; i < line; i++) {

for (j = 0; j < line - i - 1; j++){

printf(" ");

}

for (j = 0; j <= i; j++){

printf("%ld ", fact(i) / (fact(j) * fact(i - j)));

}

printf("\n");

}
return 0;

}

long fact(int num) {

long f = 1;
int i = 1;

while (i <= num) {

f = f * i;

i++;
}

return f;

}
OUTPUT

Enter the no. of lines: 4
1
1 1
1 2 1
1 3 3 1

Answer 4

thanks for the suggestion

Answer 5

What in the question is not pascal triagle.