794. Valid Tic-Tac-Toe State 有效的井字游戏

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/

题目地址: https://leetcode.com/problems/valid-tic-tac-toe-state/description/


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


  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"}.





  1. 初始的棋盘上O的个数不等于X的个数,或者O的个数不等于X-1;
  2. 棋盘上O的个数等于X - 1(轮到O下),但是O还没下棋,此时O已经赢了;
  3. 棋盘上O的个数等于X(轮到X下),但是X还没下棋,此时X已经赢了;



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



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