[LeetCode]348. Design Tic-Tac-Toe_Medium tag: array

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Assume the following rules are for the tic-tac-toe game on an n x n board between two players:

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

Implement the TicTacToe class:

• TicTacToe(int n) Initializes the object the size of the board n.
• int move(int row, int col, int player) Indicates that player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move.

Follow up:
Could you do better than O(n2) per move() operation?

Example 1:

Input
["TicTacToe", "move", "move", "move", "move", "move", "move", "move"]
[[3], [0, 0, 1], [0, 2, 2], [2, 2, 1], [1, 1, 2], [2, 0, 1], [1, 0, 2], [2, 1, 1]]
Output
[null, 0, 0, 0, 0, 0, 0, 1]

Explanation
TicTacToe ticTacToe = new TicTacToe(3);
Assume that player 1 is "X" and player 2 is "O" in the board.
ticTacToe.move(0, 0, 1); // return 0 (no one wins)
|X| | |
| | | |    // Player 1 makes a move at (0, 0).
| | | |

ticTacToe.move(0, 2, 2); // return 0 (no one wins)
|X| |O|
| | | |    // Player 2 makes a move at (0, 2).
| | | |

ticTacToe.move(2, 2, 1); // return 0 (no one wins)
|X| |O|
| | | |    // Player 1 makes a move at (2, 2).
| | |X|

ticTacToe.move(1, 1, 2); // return 0 (no one wins)
|X| |O|
| |O| |    // Player 2 makes a move at (1, 1).
| | |X|

ticTacToe.move(2, 0, 1); // return 0 (no one wins)
|X| |O|
| |O| |    // Player 1 makes a move at (2, 0).
|X| |X|

ticTacToe.move(1, 0, 2); // return 0 (no one wins)
|X| |O|
|O|O| |    // Player 2 makes a move at (1, 0).
|X| |X|

ticTacToe.move(2, 1, 1); // return 1 (player 1 wins)
|X| |O|
|O|O| |    // Player 1 makes a move at (2, 1).
|X|X|X|

Constraints:

• 2 <= n <= 100
• player is 1 or 2.
• 0 <= row, col < n
• (row, col) are unique for each different call to move.
• At most n2 calls will be made to move.

Ideas:

for each time player play, we check row, col, dig and anti dig.  T: O(n) per move

class TicTacToe:

    def __init__(self, n: int):
        """
        Initialize your data structure here.
        """
        self.n = n
        self.board= [[0] * n for _ in range(n)]
        

    def move(self, row: int, col: int, player: int) -> int:
        """
        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.
        """
        self.board[row][col] = player
        if self.checkRow(row, player) or self.checkCol(col, player) or (row == col and self.checkDig(player)) or (row == self.n - 1 - col and self.checkAntiDig(player)):
            return player
        return 0
        
    
    def checkRow(self, row, player):
        for i in range(self.n):
            if self.board[row][i] != player:
                return False
        return True
    
    def checkCol(self, col, player):
        for i in range(self.n):
            if self.board[i][col] != player:
                return False
        return True
    
    def checkDig(self, player):
        for i in range(self.n):
            if self.board[i][i] != player:
                return False
        return True
    
    def checkAntiDig(self, player):
        for i in range(self.n):
            if self.board[i][self.n - i - 1] !=player:
                return False
        return True

2. T: O(1), S: O(n), 利用row 和col来记录每行及每列每个player放的个数，如果 == self.n， 那么win。

class TicTacToe:

    def __init__(self, n: int):
        """
        Initialize your data structure here.
        """
        self.n = n
        self.board= [[0] * n for _ in range(n)]
        self.row = [[0] * n for _ in range(2)]   # 0 : player 1
        self.col = [[0] * n for _ in range(2)]   # 0 : player 1
        self.dig = [0] * 2
        self.anti = [0] * 2
        

    def move(self, row: int, col: int, player: int) -> int:
        """
        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.
        """
        self.board[row][col] = player
        self.row[player - 1][row] += 1
        self.col[player - 1][col] += 1
        if row == col:
            self.dig[player - 1] += 1
        if row == self.n - col - 1:
            self.anti[player - 1] += 1
        if self.n in [self.row[player - 1][row],  self.col[player - 1][col], self.dig[player - 1], self.anti[player - 1]]: 
            return player
        return 0

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