Check for duplication of number in ArrayList - Java

Question

I'm currently making a sudoku program, however my current code seems to fail me. The script below should put out a print "Inconsistent sudoku puzzle" if a row contains the same number several times, but sadly it doesn't.. I've tried several different attempts but no succes.

public void checkRow() {
    int count = 0;

    for(int j = 0; j < list.size(); j++) {
        for(int a = 1; a < 10; a++) {
            for (int i=0; i < list.get(j).length(); i++) {
                if(list.get(j).charAt(i) == a) {
                    count++;
                    if(count >= 2) {
                        System.out.println("Inconsistent sudoku puzzle");
                        count = 0;
                    }
                }
            }
            count = 0;
        }
    }
}

This is the collection of all my error checks:

public void errorCheck() {
    this.checkRow();
    this.checkColumn();
    this.checkBox();
}

Here i load it into my main. The code is a lot more elaborate, but these should be the sections involving the issue.

public static void main(String[] args) throws Exception  {
    Sudoku s = new Sudoku("C:\\Users\\caspe\\Downloads\\Sudoku001.sdk");
    s.printBoard();
    s.errorCheck();
    s.getNum();


    while(getNum() > 0) {
        System.out.println("Next move, please (row , column , value )");
        Scanner scanner = new Scanner(System.in);
        int row = scanner.nextInt();
        int column = scanner.nextInt() ;
        int value = scanner.nextInt();
        if (s.moves(row, column, value)); {
            s.errorCheck();
        }

        s.printBoard();
    }
}

Mark · Accepted Answer · 2018-11-04 20:42:30Z

2

The issue

You're using charAt and trying to compare the result of that to a number:

list.get(j).charAt(i) == a

However doing so you're comparing the ascii value of the character to the number.

Example:

String a = "3";
System.out.println((int) a.charAt(0)); // This prints 51

The solution

If you wanted to compare number values you'd have to do something like this:

String a = "3";
System.out.println(Character.getNumericValue(a.charAt(0))); // This prints 3

Character.getNumericValue(a.charAt(0)) returns the number value of the character.

Implementation

Implementing that into your code would look like this:

Character.getNumericValue(list.get(j).charAt(i)) == a

answered Nov 4, 2018 at 20:42

Mark

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Comments

forpas · Accepted Answer · 2018-11-04 20:43:49Z

1

This line:

if(list.get(j).charAt(i) == a)

is always false because you compare a char with an int.

Replace it with

if((list.get(j).charAt(i)-'0') == a)

list.get(j).charAt(i)-'0' gives you the numeric representation of the char

answered Nov 4, 2018 at 20:43

forpas

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Comments

Mico Heltborg · Accepted Answer · 2018-11-04 20:55:32Z

1

the problem is: 'if(list.get(j).charAt(i) == a)'

its comparing with the "a" value on the ascii table

answered Nov 4, 2018 at 20:55

Mico Heltborg

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