Question

348. Design Tic-Tac-Toe

中文文档

Description

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 the player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move, and the two players alternate in making moves. Return
  • 0 if there is no winner after the move,
  • 1 if player 1 is the winner after the move, or
  • 2 if player 2 is the winner after the move.

 

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.

 

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

Solutions

Solution 1

class TicTacToe:
  def __init__(self, n: int):
    """
    Initialize your data structure here.
    """
    self.n = n
    self.counter = [[0] * ((n << 1) + 2) for _ in range(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.
    """
    n = self.n
    self.counter[player - 1][row] += 1
    self.counter[player - 1][col + n] += 1
    if row == col:
      self.counter[player - 1][n << 1] += 1
    if row + col == n - 1:
      self.counter[player - 1][(n << 1) + 1] += 1
    if (
      self.counter[player - 1][row] == n
      or self.counter[player - 1][col + n] == n
      or self.counter[player - 1][n << 1] == n
      or self.counter[player - 1][(n << 1) + 1] == n
    ):
      return player
    return 0


# Your TicTacToe object will be instantiated and called as such:
# obj = TicTacToe(n)
# param_1 = obj.move(row,col,player)

class TicTacToe {
  private int n;
  private int[][] counter;

  /** Initialize your data structure here. */
  public TicTacToe(int n) {
    counter = new int[2][(n << 1) + 2];
    this.n = n;
  }

  /**
     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.
   */
  public int move(int row, int col, int player) {
    counter[player - 1][row] += 1;
    counter[player - 1][col + n] += 1;
    if (row == col) {
      counter[player - 1][n << 1] += 1;
    }
    if (row + col == n - 1) {
      counter[player - 1][(n << 1) + 1] += 1;
    }
    if (counter[player - 1][row] == n || counter[player - 1][col + n] == n
      || counter[player - 1][n << 1] == n || counter[player - 1][(n << 1) + 1] == n) {
      return player;
    }
    return 0;
  }
}

/**
 * Your TicTacToe object will be instantiated and called as such:
 * TicTacToe obj = new TicTacToe(n);
 * int param_1 = obj.move(row,col,player);
 */