LeetCode 1275. Find Winner on a Tic Tac Toe Game 在井字游戏中寻找获胜者 (Easy)

Tic-tac-toe is played by two players A and B on a 3 x 3 grid.
井字游戏由两名玩家A和B在3 x 3的网格上进行。

Here are the rules of Tic-Tac-Toe:
这是井字游戏的规则:

Players take turns placing characters into empty squares (" ").
玩家轮流将角色放置在空的正方形(“”)中。

The first player A always places "X" characters, while the second player B always places "O" characters.
第一个玩家A总是放置“ X”个字符,而第二个玩家B总是放置“ O”个字符。

"X" and "O" characters are always placed into empty squares, never on filled ones.
“ X”和“ O”字符始终放置在空的正方形中,决不能放在填充的正方形中。

The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
当有3个相同(非空)字符填充任何行,列或对角线时,游戏结束。

The game also ends if all squares are non-empty.
如果所有方块都不为空,则游戏也会结束。

No more moves can be played if the game is over.
如果游戏结束,将无法再进行任何移动。

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.
给定一个数组移动,其中每个元素是大小为2的另一个数组,对应于网格的行和列,它们按照A和B的播放顺序标记各自的字符。

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".
如果游戏赢家(A或B)存在,则将其返回,如果游戏以平局结束则返回“ Draw”,如果仍有动作要进行,则返回“ 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.
您可以假设移动是有效的(遵循井字游戏规则),网格最初是空的,A将首先开始游戏

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.

Solution:

class Solution:
    def tictactoe(self, moves: List[List[int]]) -> str:
        n = 3
        rows = [0] * n
        cols = [0] * n
        diag1 = diag2 = 0
        for idx, move in enumerate(moves):
            i, j = move
            sign = 1 if idx % 2 == 0 else -1
            rows[i] += sign
            cols[j] += sign
            if i == j :
                diag1 += sign
            if i + j == n-1: # reverse diag
                diag2 += sign
            if abs(rows[i]) == n or abs(cols[j]) == n or abs(diag1) == n or abs(diag2) == n:
                return "A" if sign == 1 else "B"
        return "Draw" if len(moves) == n * n else "Pending"

This solution is self-explaining. Just need to notice that the reverse diag is i + j == n-1
该解决方案是不言自明的。只需注意反向对角线的点坐标是 i + j == n-1

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