How to change an ArrayList into a 2D array in Java

Question

I am supposed to be making a program that tests a user inputed matrix is a magic square. Basically I should be putting user input into an ArrayList which is then placed into a 2D array that can then be used to calculate the sum of the rows, col, and diagonals to see if they have the same sum. This is what I have so far. I cant get the ArrayList to make a 2D array.

import java.util.*;

class Square
{
   private int[][] square;
   private ArrayList<Integer> numbers;
   public int numInput;

   public Square()
   {
      numbers = new ArrayList<Integer>(); 
      int[][] square;
      numInput = 0;
   }

   public void add(int i)
   {
      numbers.add(i);
   }
}

   public boolean isSquare()
   {
      numInput = numbers.size();
      double squared = Math.sqrt(numInput);

      if (squared != (int)squared)
      {
         System.out.println("Numbers make a square");
         return true;
      }
      else
      {
         System.out.println("Numbers do not make a square");
         return false;
      }
   }

      public String isMagicSquare()
      {

         for (int row=0; row<numInput; row++) 
         {
            for (int col=0; col<numInput; col++)
            {
               square[row][col] = number.get(col +( number.size() * row));
            }
         }
      }
}

int[][] square; declares a new array. I do not think that is what you want. You need to initialize this.square in the constructor. — OneCricketeer
– OneCricketeer, Commented Nov 2, 2016 at 17:04
If you do not need the int[][], then I made an answer on this post with just a List — OneCricketeer
– OneCricketeer, Commented Nov 2, 2016 at 17:06
Caitlyn, I'd love to match your expectations. Would you please respond to my answer and tell me if it's good or not? — xenteros
– xenteros, Commented Nov 2, 2016 at 17:32

xenteros · Accepted Answer · 2016-11-02 17:23:24Z

2

I see two cases:

User gives the size at the beginning
User doesn't.

Ad. 1.
No need to use an ArrayList. Simply read the input this way:

Scanner s = new Scanner(System.in);
int n = s.nextInt();
int[][] array = new int[n][n];
for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
        array[i][j] = s.nextInt();
    }
}

Ad. 2.

I simply scan numbers as long, as user gives numbers. Then check if he gave proper amount of numbers. Then convert to a square array of ints.

ArrayList<Integer> list = new ArrayList<>();
Scanner s = new Scanner(System.in);
while (s.hasNextInt()) {
    list.add(s.nextInt());
}
int n = list.size();
double sqrt = Math.sqrt(n);
int x = (int) sqrt;
if(Math.pow(sqrt,2) != Math.pow(x,2)) {
    //wrong input - it wasn't a square
}
int[][] array = new int[x][x];
int index = 0;
for (int i = 0; i < x; i++) {
    for (int j = 0; j < x; j++) {
        array[i][j] = array.get(index++);
    }
}

Obviously you need to take care about error handling. If you have further questions, ask in comments. I'll update my answer if you're interested.

edited Nov 2, 2016 at 17:23

answered Nov 2, 2016 at 17:08

xenteros

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1 Comment

xenteros Over a year ago

@CaitlynDaly on Stack Overflow instead of writing 'thanks' we upvote and mark the answer as accepted. Just in case, you wanna join the community I strongly recommend reading How to Ask and minimal reproducible example. Especially the second one will teach you how to ask questions in a way, that I wouldn't need to guess which case we're talking about. Good luck on Stack Overflow!

Saravana · Accepted Answer · 2016-11-02 17:39:01Z

0

There is a typo in identifying the perfect square

it should be

if (squared == (int) squared) return true;

You can initialize and fill 2D array if its a perfect square

public String isMagicSquare() {
    if (isSquare()) {
        int size = (int) Math.sqrt(numbers.size());
        this.square = new int[size][size];
        for (int i = 0; i < numbers.size(); i++) {
            square[i / size][i % size] = numbers.get(i);
        }
        return Arrays.deepToString(square); // do other op on the array and return appropriate String
    } else {
        return null; 
    }
}

answered Nov 2, 2016 at 17:39

Saravana

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