1275. Find Winner on a Tic Tac Toe Game-找出井字棋的获胜者

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.

Example 1:
Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
Output: "A"
Explanation: "A" wins, he always plays first.

Example 2:
Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: "B" wins.

Example 3:
Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
Output: "Draw"
Explanation: The game ends in a draw since there are no moves to make.

Example 4:
Input: moves = [[0,0],[1,1]]
Output: "Pending"
Explanation: The game has not finished yet.

Constraints:

• 1 <= moves.length <= 9
• moves[i].length == 2
• 0 <= moves[i][j] <= 2
• There are no repeated elements on moves.
• moves follow the rules of tic tac toe.

Solution in python:

class Solution:
    def tictactoe(self, moves: List[List[int]]) -> str:
        flag = 1
        char = ['X', 'O']
        name = ['A', 'B']
        broad = [[' ' for i in range(3)] for j in range(3)]
        for item in moves:
            if flag:
                broad[item[0]][item[1]] = 'X'
            else:
                broad[item[0]][item[1]] = 'O'
            flag = 1-flag
        s = None
        for i in range(3):
            if broad[i][0] == broad[i][1] == broad[i][2]:
                s = broad[i][0]
                break
            if broad[0][i] == broad[1][i] == broad[2][i]:
                s = broad[0][i]
                break
        if broad[0][0] == broad[1][1] == broad[2][2]:
            s = broad[0][0]
        if broad[2][0] == broad[1][1] == broad[0][2]:
            s = broad[2][0]
        if s and (s == 'X' or s == 'O'):
            return name[char.index(s)]
        else:
            if len(moves) < 9:
                return "Pending"
            else:
                return "Draw"