LeetCode 348. Design Tic-Tac-Toe（井字棋）

原题网址：https://leetcode.com/problems/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?

Hint:

1. Could you trade extra space such that move() operation can be done in O(1)?
2. You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

方法：类似八皇后问题，记录各行各列以及对角线的落子情况。

public class TicTacToe {
    private int[][] rows = new int[2][];
    private int[][] cols = new int[2][];
    private int[] diag = new int[2];
    private int[] adiag = new int[2];
    private int n;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        for(int i=0; i

简化：

public class TicTacToe {
    private int[][] rows;
    private int[][] cols;
    private int[] diag = new int[2];
    private int[] adiag = new int[2];
    private int size;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        size = n;
        rows = new int[n][2];
        cols = new int[n][2];
    }
    
    /** 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) {
        rows[row][player-1] ++;
        if (rows[row][player-1] == size) return player;
        cols[col][player-1] ++;
        if (cols[col][player-1] == size) return player;
        if (row == col) {
            diag[player-1] ++;
            if (diag[player-1] == size) return player;
        }
        if (row+col==size-1) {
            adiag[player-1] ++;
            if (adiag[player-1] == size) 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);
 */