Leetcode 794. Valid Tic-Tac-Toe State

文章作者：Tyan
博客：noahsnail.com  |  CSDN  |  简书

1. Description

2. Solution

解析：Version 1，判断游戏合不合法，主要分为下面几个方面：

1. 由于X先放，轮流放置，因此X的数量永远大于等于O的数量。
2. 由于是轮流放置，因此二者的数量差值最大为1。
3. 当X先结束游戏时，此时X的数量等于O的数量加1。
4. 当O先结束游戏时，此时X的数量等于O的数量。
   根据上述条件依次判断即可。

• Version 1

class Solution:
    def validTicTacToe(self, board: List[str]) -> bool:
        x_count = 0
        o_count = 0
        for line in board:
            for ch in line:
                if ch == 'X':
                    x_count += 1
                elif ch == 'O':
                    o_count += 1
        if o_count > x_count or x_count > o_count + 1:
            return False
        if x_count == o_count and self.isGameOver(board, 'X'):
            return False
        if x_count == o_count + 1 and self.isGameOver(board, 'O'):
            return False

        return True


    def isGameOver(self, board: List[str], ch: str) -> bool:
        # Check rows
        for i in range(3):
            if board[i][0] == ch and board[i][0] == board[i][1] and board[i][1] == board[i][2]:
                return True

        # Check columns
        for i in range(3):
            if board[0][i] == ch and board[0][i] == board[1][i] and board[1][i] == board[2][i]:
                return True

        # Check diagonals
        if board[1][1] == ch and ((board[0][0] == board[1][1] and board[1][1] == board[2][2]) or
            (board[0][2] == board[1][1] and board[1][1] == board[2][0])):
                return True
        return False

Reference

1. https://leetcode.com/problems/valid-tic-tac-toe-state/