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Java Array Exercises: Print the specified grid

Java Array: Exercise-3 with Solution

Write a Java program to print the following grid.

Sample Solution:

Java Code:

public class Main {
public static void main(String[] args) {   
 int [][]a = new int[10][10];    
 for(int i = 0; i < 10; i++)
   {
      for(int j = 0; j < 10; j++)
      {
         System.out.printf("%2d ", a[i][j]);
      }
      System.out.println();
   }
}
}

Sample Output: 

0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0 
 0  0  0  0  0  0  0  0  0  0

Flowchart:

Flowchart: Java exercises: Print the specified grid

Java Code Editor:

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Java: Tips of the Day

Different between parseInt() and valueOf() in java?

Well, the API for Integer.valueOf(String) does indeed say that the String is interpreted exactly as if it were given to Integer.parseInt(String). However, valueOf(String) returns a new Integer() object whereas parseInt(String) returns a primitive int.

If you want to enjoy the potential caching benefits of Integer.valueOf(int), you could also use this eyesore:

Integer k = Integer.valueOf(Integer.parseInt("123"))

Now, if what you want is the object and not the primitive, then using valueOf(String) may be more attractive than making a new object out of parseInt(String) because the former is consistently present across Integer, Long, Double, etc.

Ref: https://bit.ly/3vRuIPY