[LeetCode]Valid Tic-Tac-Toe State

题目描述：

LeetCode 794. Valid Tic-Tac-Toe State

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: falseExplanation: The first player always plays "X".Example 2:Input: board = ["XOX", " X ", "   "]Output: falseExplanation: Players take turns making moves.Example 3:Input: board = ["XXX", "   ", "OOO"]Output: falseExample 4:Input: board = ["XOX", "O O", "XOX"]Output: true

Note:

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

题目大意：

给定一个井字棋的棋盘，判断棋盘状态是否合法。

'X'先'O'后，当三子连珠时（横、竖、对角线）游戏结束。

解题思路：

计算棋盘上'X'和'O'的个数，记为nx, no

分别判断'X'和'O'是否存在三子连珠，记为wx, wo

若wx为真，则nx == no + 1并且wo为假

若wo为真，则nx == no

否则，nx - 1 <= no <= nx

Python代码：

class Solution(object):
    def validTicTacToe(self, board):
        """
        :type board: List[str]
        :rtype: bool
        """
        nx = ''.join(board).count('X')
        no = ''.join(board).count('O')
        wx, wo = self.isWin(board, 'X'), self.isWin(board, 'O')
        if wx: return nx == no + 1 and not wo
        if wo: return nx == no
        return nx - 1 <= no <= nx

    def isWin(self, board, pc):
        if any(r == pc * 3 for r in board): return True
        if any(c == pc * 3 for c in zip(*board)): return True
        if board[0][0] == board[1][1] == board[2][2] == pc: return True
        if board[0][2] == board[1][1] == board[2][0] == pc: return True
        return False