How can I check if the first row of a 2D array with no fixed dimension all have the same value?-CodePudding

I'm making a tic tac toe game in Java with a customizable dimension feature (user can choose to play 3x3, 4x4, 5x5, etc.) and was working on the logic regarding finding a winner. Currently I'm trying to figure out checking for wins horizontally.

I've had the idea to make a nested for-loop to check the 2d array that hosts the board, but am not sure how to execute this. The problem with this code:

for (int i = 0; i < dimension; i  ) {
        if (board[i][0] == board[i][1] && board[i][1] == board[i][2] && board[i][0] != '-') {
                // you won!
        }
}

...Is that its logic is fixed for a 3x3 game, not for any other dimensions. I only know how to add values into 2d arrays, so how can I check if these values are equal? Thank you in advance.

CodePudding user response：

I'm not sure about the rules for winning tic tac toe in higher dimensions, so let's say you have to fill the whole row/column to win.

Divide the if in two parts: The check for the first chararacter and the comparisons. Then use a second for-loop for the comparisons, like this:

for (int i = 0; i < dimension; i  ) { // iterate rows
    // check for first character
    if (board[i][0] == '-') {    // if wrong character...
        continue;                // ... check next row
    }

    boolean won = true;

    for (int j = 0; j < dimension - 1; j  ) { // iterate columns
        if (board[i][j] != board[i][j 1]) {   // if other character...
            won = false;                      // ...not winnable with this column and...
            break;                            // ...stop iteration of columns
        }
    }

    if (won) {
        // you won!
    }
}

If you win with less crosses 'X' in all dimensions, you would have to add a third loop to go through the possible start points or you could count the amount of crosses in the column and reset the number if there is an 'O'.