348. Design Tic-Tac-Toe

348. 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:

1. A move is guaranteed to be valid and is placed on an empty block.
2. Once a winning condition is reached, no more moves is allowed.
3. 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?

-------------------
Solution #1
开一个2d array

class TicTacToe {

    private int[][] grid;
    private int n;
    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        this.n = n;
        grid = new int[n][n];
    }
    
    /** Player {player} makes a move at ({row}, {col}).
        @param row The row of the board.
        @param col The column of the board.
        @param player The player, can be either 1 or 2.
        @return The current winning condition, can be either:
                0: No one wins.
                1: Player 1 wins.
                2: Player 2 wins. */
    public int move(int row, int col, int player) {
        grid[row][col] = player;
        
        int win = player;
        for (int i = 0; i < n; i++) {
            if (grid[row][i] != player) {
                win = 0;
                break;
            }
        }
        
        if (win == player) return player;
        win = player;
        
        for (int i = 0; i < n; i++) {
            if (grid[i][col] != player) {
                win = 0;
                break;
            }
        }
        
        if (win == player) return player;
        
        if (row == col) {
            win = player;
            for (int i = 0; i < n; i++) {
                if (grid[i][i] != player) {
                    win = 0;
                    break;
                }
            }
            if (win == player) return player;
        }
        
        
        if (row + col == n - 1) {
            win = player;
            for (int i = 0; i < n; i++) {
                if (grid[i][n - 1 - i] != player) {
                    win = 0;
                    break;
                }
            }
            if (win == player) return player;
        }
        
        return 0;
    }    
}

/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe obj = new TicTacToe(n);
 * int param_1 = obj.move(row,col,player);
 */

Solution #2，对1的优化
用2个array分别记录每一行和列，2个值来记录对角线

class TicTacToe {
    
    private int[] rows;
    private int[] cols;
    private int diag;
    private int antiDiag;
    private int n;
    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        this.n = n;
        rows = new int[n];
        cols = new int[n];
        diag = 0;
        antiDiag = 0;
    }
    
    /** Player {player} makes a move at ({row}, {col}).
        @param row The row of the board.
        @param col The column of the board.
        @param player The player, can be either 1 or 2.
        @return The current winning condition, can be either:
                0: No one wins.
                1: Player 1 wins.
                2: Player 2 wins. */
    public int move(int row, int col, int player) {
        int mv = player == 1? 1 : -1;
        rows[row] += mv;
        cols[col] += mv;
        
        if (row == col) diag += mv;
        if (row + col == n - 1) antiDiag += mv;
        
        if (Math.abs(rows[row]) == n 
            || Math.abs(cols[col]) == n
            || Math.abs(diag) == n
            || Math.abs(antiDiag) == n) {
            
            return player;
        }
        
        return 0;
    }
}

/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe obj = new TicTacToe(n);
 * int param_1 = obj.move(row,col,player);
 */