題目描述：

A Tic-Tac-Toe board is given as a string array board. Return True if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array, and consists of characters " ", “X”, and “O”. The " " character represents an empty square.

Here are the rules of Tic-Tac-Toe:

Players take turns placing characters into empty squares (" "). The first player always places “X” characters, while the second player always places “O” characters. “X” and “O” characters are always placed into empty squares, never filled ones. The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal. The game also ends if all squares are non-empty. No more moves can be played if the game is over.

Example 1:

Input: board = ["O  ", "   ", "   "]
Output: false
Explanation: The first player always plays "X".

Example 2:

Input: board = ["XOX", " X ", "   "]
Output: false
Explanation: Players take turns making moves.

Example 3:

Input: board = ["XXX", "   ", "OOO"]
Output: false

Example 4:

Input: board = ["XOX", "O O", "XOX"]
Output: true
 

Note:

1. board is a length-3 array of strings, where each string board[i] has length 3.
2. Each board[i][j] is a character in the set {" ", “X”, “O”}.

題目大意

判斷一個棋盤是不是有效的井字棋的狀態。

解題方法

判斷是否是個合法的狀態，我只需要排除不合法的狀態就好了。不合法的狀態分為三種情況：

1. 初始的棋盤上O的個數不等於X的個數，或者O的個數不等於X-1；
2. 棋盤上O的個數等於X - 1（輪到O下），但是O還沒下棋，此時O已經贏了；
3. 棋盤上O的個數等於X（輪到X下），但是X還沒下棋，此時X已經贏了；

除此之外，就能下棋，或者棋局已經結束。

時間複雜度是O(N^2)，空間複雜度是O(1)。N是棋盤變長。

class Solution:
    def validTicTacToe(self, board):
        """
        :type board: List[str]
        :rtype: bool
        """
        xCount, oCount = 0, 0
        for i in range(3):
            for j in range(3):
                if board[i][j] == 'O':
                    oCount += 1
                elif board[i][j] == 'X':
                    xCount += 1
        if oCount != xCount and oCount != xCount - 1: return False
        if oCount != xCount and self.win(board, 'O'): return False
        if oCount != xCount - 1 and self.win(board, 'X'): return False
        return True
        
    def win(self, board, P):
        # board is list[str]
        # P is 'X' or 'O' for two players
        for j in range(3):
            if all(board[i][j] == P for i in range(3)): return True
            if all(board[j][i] == P for i in range(3)): return True
        if board[0][0] == board[1][1] == board[2][2] == P: return True
        if board[0][2] == board[1][1] == board[2][0] == P: return True
        return False

參考資料：

日期

2018 年 10 月 14 日 —— 美好的週一怎麼會出現霧霾呢？