[Swift]LeetCode348. 设计井字棋游戏 $ Design Tic-Tac-Toe

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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?

Hint:

Could you trade extra space such that move() operation can be done in O(1)?
You need two arrays: int rows[n], int cols[n], plus two variables: diagonal, anti_diagonal.

---

设计一个井字游戏，在N x N网格上的两个玩家之间进行。 

您可以假定以下规则： 

移动被保证是有效的，并且被放置在一个空块上。

一旦达到获胜条件，就不允许再移动。

在横排、竖排或斜排中成功放置N个标记的玩家获胜。

例子：

假设n=3，假设玩家1是“x”，玩家2是“o”。

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|

跟进：

每个move()操作可以做得比O(n2) 更好吗？

 提示： 

您是否可以交换额外的空间，以便在O(1)中执行move()操作？

您需要两个数组：int rows[n]、int cols[n]，加上两个变量：对角线、反斜线。

---

Solution:

class TicTacToe {
    var diag:Int = 0
    var rev_diag:Int = 0
    var N:Int = 0
    var rows:[Int] = [Int]()
    var cols:[Int] = [Int]()
    
    init(_ n:Int)
    {
        self.rows = [Int](repeating:0,count:n)
        self.cols = [Int](repeating:0,count:n)
        self.N = n
    }
    
    func move(_ row:Int,_ col:Int,_ player:Int) -> Int
    {
        var add:Int = player == 1 ? 1 : -1
        rows[row] += add
        cols[col] += add
        diag += (row == col ? add : 0)
        rev_diag += (row == N - col - 1 ? add : 0)
        if abs(rows[row]) == N || abs(cols[col]) == N || abs(diag) == N || abs(rev_diag) == N
        {
            return player
        }
        return 0
    }    
}

点击：Playground测试

var toe:TicTacToe = TicTacToe(3)

//toe.move(0, 0, 1); -> Returns 0 (no one wins)
print(toe.move(0, 0, 1))
//Print 0

//toe.move(0, 2, 2); -> Returns 0 (no one wins)
print(toe.move(0, 2, 2))
//Print 0

//toe.move(2, 2, 1); -> Returns 0 (no one wins)
print(toe.move(2, 2, 1))
//Print 0

//toe.move(1, 1, 2); -> Returns 0 (no one wins)
print(toe.move(1, 1, 2))
//Print 0

//toe.move(2, 0, 1); -> Returns 0 (no one wins)
print(toe.move(2, 0, 1))
//Print 0

//toe.move(1, 0, 2); -> Returns 0 (no one wins)
print(toe.move(1, 0, 2))
//Print 0

//toe.move(2, 1, 1); -> Returns 1 (player 1 wins)
print(toe.move(2, 1, 1))
//Print 1