题目如下：

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.

"X  "    "X  "    "X  "    "X  "    "X  "

"   " -> "   " -> " X " -> " X " -> " X "

"   "    "O  "    "O  "    "OO "    "OOX"

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]

Output: "B"

Explanation: "B" wins.

"X  "    "X  "    "XX "    "XXO"    "XXO"    "XXO"

"   " -> " O " -> " O " -> " O " -> "XO " -> "XO "

"   "    "   "    "   "    "   "    "   "    "O  "

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.

"XXO"

"OOX"

"XOX"

Example 4:

Input: moves = [[0,0],[1,1]]

Output: "Pending"

Explanation: The game has not finished yet.

"X  "

" O "

"   "

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.

解题思路：把moves都下完后，判断当前棋盘上的是否存在三连即可。

代码如下：

class Solution(object):

    def tictactoe(self, moves):

        """

        :type moves: List[List[int]]

        :rtype: str

        """

        grid = [[0] * 3 for _ in range(3)]

        for i in range(len(moves)):

            x, y = moves[i]

            grid[x][y] = 1 if i%2 == 0 else -1

        if sum(grid[0]) == 3 or sum(grid[1]) == 3 or sum(grid[2]) == 3:

            return "A"

        elif sum(grid[0]) == -3 or sum(grid[1]) == -3 or sum(grid[2]) == -3:

            return "B"

        elif (grid[0][0] + grid[1][0] + grid[2][0]) == 3 or (grid[0][1] + grid[1][1] + grid[2][1]) == 3 or \

                (grid[0][2] + grid[1][2] + grid[2][2]) == 3:

            return "A"

        elif (grid[0][0] + grid[1][0] + grid[2][0]) == -3 or (grid[0][1] + grid[1][1] + grid[2][1]) == -3 or \

                (grid[0][2] + grid[1][2] + grid[2][2]) == -3:

            return "B"

        elif (grid[0][0] + grid[1][1] + grid[2][2] == 3) or (grid[0][2] + grid[1][1] + grid[2][0] == 3):

            return "A"

        elif (grid[0][0] + grid[1][1] + grid[2][2] == -3) or (grid[0][2] + grid[1][1] + grid[2][0] == -3):

            return "B"

        elif len(moves) == 9:

            return "Draw"

        return "Pending"