1275. Find Winner on a Tic Tac Toe Game

Tic-tac-toe is played by two players A and B on a 3 x 3 grid.

Here are the rules of Tic-Tac-Toe:

• Players take turns placing characters into empty squares (" ").
• The first player A always places "X" characters, while the second player B always places "O" characters.
• "X" and "O" characters are always placed into empty squares, never on 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.

Given an array moves where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which A and B play.

Return the winner of the game if it exists (A or B), in case the game ends in a draw return "Draw", if there are still movements to play return "Pending".

You can assume that moves is valid (It follows the rules of Tic-Tac-Toe), the grid is initially empty and A will play first.

一行或一列或者对角线如果全是一个人棋，那么这个人就赢了。注意还有两种状态，和棋和正在下，"Draw" and "Pending"

我感觉操作二维数组的写法有点麻烦，所以转换成一维的来判断。

class Solution(object):
    def tictactoe(self, moves):
        """
        :type moves: List[List[int]]
        :rtype: str
        """
        # 0 1 2
        # 3 4 5
        # 6 7 8
        candidates = [
            [0, 1, 2],
            [3, 4, 5],
            [6, 7, 8],
            [0, 3, 6],
            [1, 4, 7],
            [2, 5, 8],
            [0, 4, 8],
            [2, 4, 6]
        ]
        grid = [-1] * 9
        player = 0
        for move in moves:
            grid[move[0] * 3 + move[1]] = player
            player += 1
            player %= 2
        for candidate in candidates:
            flag = True
            for i in range(1, len(candidate), 1):
                if grid[candidate[i]] != grid[candidate[i - 1]] or grid[candidate[i]] == -1:
                    flag = False
                    break
            if flag:
                return 'A' if grid[candidate[0]] == 0 else 'B'
            
        draw = 0
        for i in range(9):
            if grid[i] != -1:
                draw += 1
        if draw == 9:
            return 'Draw'
        else:
            return 'Pending'