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Lc0348

05 August 2008

348 Design Tic-Tac-Toe

题目

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:

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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 

1

move()
operation?

分析

1) 绘制棋盘，判断当前落子所在行与列是否都是该棋手的子，以及如果落子在对角线上,则同时需要判断对角线上的子是否都是该棋手的子。

2) 不需要复现棋盘，只需要计算每个player在各条线是否放了n个棋子就能知道是否获胜。 不用二维数组，近一步减小空间消耗，+1计算player1的棋子数，-1计算player2的棋子数，只要有一方达到边界就算赢（n-1，player1赢，1-n，player2赢）。时间O(1)，空间O(n)。

代码

方法1

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class TicTacToe {
    private int[][] scope;

    public TicTacToe(int n) {
        scope = new int[n][n];
    }

    public int move(int row, int col, int player) {
        scope[row][col] = player;
        // 横
        boolean win = true;
        for (int item : scope[row]) {
            if (item != player) {
                win = false;
                break;
            }
        }
        if (win) {
            return player;
        }
        // 竖
        win = true;
        for (int i = 0; i < scope.length; i++) {
            if (scope[i][col] != player) {
                win = false;
                break;
            }
        }
        if (win) {
            return player;
        }
        // 左对角
        win = true;
        for (int i = 0; i < scope.length; i++) {
            if (scope[i][i] != player) {
                win = false;
                break;
            }
        }
        if (win) {
            return player;
        }
        // 有对角
        win = true;
        for (int i = 0; i < scope.length; i++) {
            if (scope[i][scope.length - 1 - i] != player) {
                win = false;
                break;
            }
        }
        if (win) {
            return player;
        } else {
            return 0;
        }
    }
}

方法2

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class TicTacToe {
    int[] rows, cols, dig;
    int n;
    public TicTacToe(int n) {
        rows = new int[n];
        cols = new int[n];
        dig = new int[2];
        this.n = n;
    }
    public int move(int row, int col, int player) {
        return ((player == 1 && rows[row]++ == n-1 | cols[col]++ == n-1 | (row == col && dig[0]++ == n-1) | (row + col == n-1 && dig[1]++ == n-1)) || (player == 2 && rows[row]-- == 1-n | cols[col]-- == 1-n | (row == col && dig[0]-- == 1-n) | (row + col == n-1 && dig[1]-- == 1-n))) ? player : 0;
    }
}
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