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Design Tic-Tac-Toe

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题目(LeetCode #348)

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.

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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?

题解

可使用二维数组分别保存两位选手的每一行、每一列和两条对角线上已放置的棋子数。由于每一步都是有效的且放置于空位，因此无需考虑每一行具体如何放置，记录数量即可，当行、列或对角线上的棋子数达到 n 时即有一方获胜。

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class TicTacToe {
    private final int[][] rows;
    private final int[][] cols;
    private final int[][] diags;
    private final int n;

    public TicTacToe(int n) {
        rows = new int[2][n];
        cols = new int[2][n];
        diags = new int[2][2];
        this.n = n;
    }

    public int move(int row, int col, int player) {
        int playerIdx = player - 1;

        rows[playerIdx][row]++;
        if (rows[playerIdx][row] == n) return player;

        cols[playerIdx][col]++;
        if (cols[playerIdx][col] == n) return player;

        if (row == col) {
            diags[playerIdx][0]++;
            if (diags[playerIdx][0] == n) return player;
        }

        if (row + col == n - 1) {
            diags[playerIdx][1]++;
            if (diags[playerIdx][1] == n) return player;
        }

        return 0;
    }
}