[LeetCode] 348. Design Tic-Tac-Toe 设计井字棋游戏

Design a Tic-tac-toe game that is played between two players on a n x n grid.

You may assume the following rules:

A move is guaranteed to be valid and is placed on an empty block.
Once a winning condition is reached, no more moves is allowed.
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(n^2) per move() operation?

Hint:

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

解法1: 暴力解法，每走一步，对所走点的水平，竖直，对角线，反对角线进行检查是否满足条件。

解法2: 根据提示，分别建立水平，竖直两个数组，以及对角线，反对角线两个变量。每走一步分别对这几个进行判断，一个玩家加1，一个玩家-1，如果下棋子的点的水平或者垂直数组里的元素的值等于n, 或者对角线的值的绝对值等于n，那么就返回此时下棋子的选手赢。

Java:

public class TicTacToe {

    int[][] matrix;

    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        matrix = 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) {
        matrix[row][col]=player;

        //check row
        boolean win=true;
        for(int i=0; i<matrix.length; i++){
            if(matrix[row][i]!=player){
                win=false;
                break;
            }
        }

        if(win) return player;

        //check column
        win=true;
        for(int i=0; i<matrix.length; i++){
            if(matrix[i][col]!=player){
                win=false;
                break;
            }
        }

        if(win) return player;

        //check back diagonal
        win=true;
        for(int i=0; i<matrix.length; i++){
            if(matrix[i][i]!=player){
                win=false;
                break;
            }
        }

        if(win) return player;

        //check forward diagonal
        win=true;
        for(int i=0; i<matrix.length; i++){
            if(matrix[i][matrix.length-i-1]!=player){
                win=false;
                break;
            }
        }

        if(win) return player;

        return 0;
    }
}

Java:

public class TicTacToe {
    int[] rows;
    int[] cols;
    int dc1;
    int dc2;
    int n;
    /** Initialize your data structure here. */
    public TicTacToe(int n) {
        this.n=n;
        this.rows=new int[n];
        this.cols=new int[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 val = (player==1?1:-1);

        rows[row]+=val;
        cols[col]+=val;

        if(row==col){
            dc1+=val;
        }
        if(col==n-row-1){
            dc2+=val;
        }

        if(Math.abs(rows[row])==n
        || Math.abs(cols[col])==n
        || Math.abs(dc1)==n
        || Math.abs(dc2)==n){
            return player;
        }

        return 0;
    }
}　　

Python:

class TicTacToe(object):

    def __init__(self, n):
        """
        Initialize your data structure here.
        :type n: int
        """
        self.__size = n
        self.__rows = [[0, 0] for _ in xrange(n)]
        self.__cols = [[0, 0] for _ in xrange(n)]
        self.__diagonal = [0, 0]
        self.__anti_diagonal = [0, 0]

    def move(self, row, col, player):
        """
        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.
        :type row: int
        :type col: int
        :type player: int
        :rtype: int
        """
        i = player - 1
        self.__rows[row][i] += 1
        self.__cols[col][i] += 1
        if row == col:
            self.__diagonal[i] += 1
        if col == len(self.__rows) - row - 1:
            self.__anti_diagonal[i] += 1
        if any(self.__rows[row][i] == self.__size,
               self.__cols[col][i] == self.__size,
               self.__diagonal[i] == self.__size,
               self.__anti_diagonal[i] == self.__size):
            return player

        return 0

C++:

class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n) {
        board.resize(n, vector<int>(n, 0));
    }

    int move(int row, int col, int player) {
        board[row][col] = player;
        int i = 0, j = 0, n = board.size();
        for (j = 1; j < n; ++j) {
            if (board[row][j] != board[row][j - 1]) break;
        }
        if (j == n) return player;
        for (i = 1; i < n; ++i) {
            if (board[i][col] != board[i - 1][col]) break;
        }
        if (i == n) return player;
        if (row == col) {
            for (i = 1; i < n; ++i) {
                if (board[i][i] != board[i - 1][i - 1]) break;
            }
            if (i == n) return player;
        }
        if (row + col == n - 1) {
            for (i = 1; i < n; ++i) {
                if (board[n - i - 1][i] != board[n - i][i - 1]) break;
            }
            if (i == n) return player;
        }
        return 0;
    }

private:
    vector<vector<int>> board;
};

C++:

class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n): rows(n), cols(n), N(n), diag(0), rev_diag(0) {}

    int move(int row, int col, int player) {
        int add = player == 1 ? 1 : -1;
        rows[row] += add;
        cols[col] += add;
        diag += (row == col ? add : 0);
        rev_diag += (row == N - col - 1 ? add : 0);
        return (abs(rows[row]) == N || abs(cols[col]) == N || abs(diag) == N || abs(rev_diag) == N) ? player : 0;
    }

private:
    vector<int> rows, cols;
    int diag, rev_diag, N;
};

　　

　　

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