Design Tic Tac Toe

/**
Design Tic-Tac-Toe
Design a Tic-tac-toe game that is played between two players on a n x n grid.
You may assume the following rules:
A move is guaranteed to be valid and is placed on an empty block.
Once a winning condition is reached, no more moves is allowed.
A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.
Example:
Given n = 3, assume that player 1 is "X" and player 2 is "O" in the board.

TicTacToe toe = new TicTacToe(3);

toe.move(0, 0, 1); -> Returns 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

toe.move(0, 2, 2); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

toe.move(2, 2, 1); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

toe.move(1, 1, 2); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

toe.move(2, 0, 1); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

toe.move(1, 0, 2); -> Returns 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|
Follow up: Could you do better than O(n2) per move() operation?
Hint:
Could you trade extra space such that move() operation can be done in O(1)? You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.
*/

/**
Soution:

The key observation is that in order to win Tic-Tac-Toe you must have the entire row or column. Thus, we don't need to keep track of an entire n^2 board. We only need to keep a count for each row and column. If at any time a row or column matches the size of the board then that player has won.
To keep track of which player, I add one for Player1 and -1 for Player2. There are two additional variables to keep track of the count of the diagonals. Each time a player places a piece we just need to check the count of that row, column, diagonal and anti-diagonal.
*/

public class DesignTicTacToe{

int[] rows, cols;
int diag, antidiag;

public DesignTicTacToe(int n){
rows = new int[n];
cols = new int[n];
}

/**
we indicate one player with 1 and another with -1
*/
int move(int row, int col, int player){
int size = rows.length;
int val = player==1?:1:-1;
rows[row]+=val;
cols[col]+=val;
if(row==col)
diag+=val;
if(row+col == rows.length)
antidiag+=val;
if(Math.abs(rows[row])==size || Math.abs(cols[col])==size || Math.abs(diag)==size || Math.abs(antidiag)==size){
return player;
}
return 0;
}
}