0348._Design_Tic-Tac-Toe.md

348. Design Tic-Tac-Toe

难度: Medium

刷题内容

原题连接

• https://leetcode.com/problems/design-tic-tac-toe/

内容描述

Design a Tic-tac-toe game that is played between two players on a n x n grid.

You may assume the following rules:

A move is guaranteed to be valid and is placed on an empty block.
Once a winning condition is reached, no more moves is allowed.
A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.
Example:
Given n = 3, assume that player 1 is "X" and player 2 is "O" in the board.

TicTacToe toe = new TicTacToe(3);

toe.move(0, 0, 1); -> Returns 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

toe.move(0, 2, 2); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

toe.move(2, 2, 1); -> Returns 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

toe.move(1, 1, 2); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

toe.move(2, 0, 1); -> Returns 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

toe.move(1, 0, 2); -> Returns 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|
Follow up:
Could you do better than O(n2) per move() operation?

解题方案

思路 1 ******- 时间复杂度: O(N) per move - 空间复杂度: O(N^2)

只需要看move的那一行和那一列是否满足win的条件即可，另外还需要考虑move的那一个点是否在主对角线和副对角线上，如果在也要考虑两条对角线是否满足win的条件

这样的做法已经满足了follow up

beats 18.55%

class TicTacToe:

    def __init__(self, n):
        """
        Initialize your data structure here.
        :type n: int
        """
        self.board = [[0] * n for i in range(n)]
        self.n = n
        
    def move(self, row, col, player):
        """
        Player {player} makes a move at ({row}, {col}).
        @param row The row of the board.
        @param col The column of the board.
        @param player The player, can be either 1 or 2.
        @return The current winning condition, can be either:
                0: No one wins.
                1: Player 1 wins.
                2: Player 2 wins.
        :type row: int
        :type col: int
        :type player: int
        :rtype: int
        """
        self.board[row][col] = player
        if all(self.board[row][i] == player for i in range(self.n)): return player
        if all(self.board[i][col] == player for i in range(self.n)): return player
        if row == col:
            if all(self.board[i][i] == player for i in range(self.n)):
                return player
        if row + col == self.n - 1:
            if all(self.board[i][self.n-1-i] == player for i in range(self.n)):
                return player
        return 0