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Print number star pattern type 1
Take input from user i.e number of lines required (N value). Take a result variable (say ‘a’) and initialize it with 1.Take two loops one for each line (say ‘i’) and other for each digit in a particular line (say ‘j’).Here ‘i’ loop is used to access each line from 1 to N and ‘j’ loop is used to print values in each line. For example in line 2 the value of i=2 and for contents in second line (i.e 5*6*7*8) the value for j for digit 5 is j=1 and for digit 6 is j=2.Print ‘a’ value along with * and post increment it until the j loop reaches a value less than n After the j loop is finished, print the ‘a’ value without * and post increment it and go to next line.Repeat the ‘i’ loop until it reaches ‘N’ lines.


Algorithm
- Take input from user i.e number of lines required (N value).
- Take a result variable (say ‘a’) and initialize it with 1.
- Take two loops one for each line (say ‘i’) and other for each digit in a particular line (say ‘j’).
- Here ‘i’ loop is used to access each line from 1 to N and ‘j’ loop is used to print values in each line. For example in line 2 the value of i=2 and for contents in second line (i.e 5*6*7*8) the value for j for digit 5 is j=1 and for digit 6 is j=2.
- Print ‘a’ value along with * and post increment it until the j loop reaches a value less than n.
- After the j loop is finished, print the ‘a’ value without * and post increment it and go to next line.
- Repeat the ‘i’ loop until it reaches ‘N’ lines.
Code in Java
import java.io.*; classPrepInsta{ public static void main(String[] args) throws IOException { BufferedReader br = new BufferedReader(new InputStreamReader(System.in)); int n, i, j, a = 1; System.out.print("Enter the number of Lines:"); n = Integer.parseInt(br.readLine()); for (i = 1; i <= n; i++){ for (j = 1; j < n; j++) { System.out.print((a++) + "*"); } System.out.println(a++); } } } This code is contributed by Shubham Nigam (Prepinsta Placement Cell Student)
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