348 Design Tic-Tac-Toe

348. Design Tic-Tac-Toe

1. Question

Design a Tic-tac-toe game that is played between two players on anxngrid.

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|


|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) permove()operation?

2. Implementation

class TicTacToe {
    int[] rows;
    int[] cols;
    int diagnol;
    int antiDiagnol;
    int n;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        rows = new int[n];
        cols = new int[n];
        this.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) {
        int toAdd = player == 1 ? 1 : -1;

        rows[row] += toAdd;
        cols[col] += toAdd;
        if (row == col) {
            diagnol += toAdd;
        }

        if (col == n - row - 1) {
            antiDiagnol += toAdd;
        }

        if (Math.abs(rows[row]) == n || Math.abs(cols[col]) == n
            || Math.abs(diagnol) == n || Math.abs(antiDiagnol) == 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);
 */

3. Time & Space Complexity

move(): 时间复杂度O(1), 空间复杂度O(n)