[LeetCode] 348. Design Tic-Tac-Toe 設計井字棋游戲

 

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

You may assume the following rules:

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 is allowed.
3. 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()

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.

 

CareerCup上的原題，請參見我之前的博客17.2 Tic Tac Toe。我們首先來O(n²)的解法，這種方法的思路很straightforward，就是建立一個nxn大小的board，其中0表示該位置沒有棋子，1表示玩家1放的子，2表示玩家2。那么棋盤上每增加一個子，我們都掃描當前行列，對角線，和逆對角線，看看是否有三子相連的情況，有的話則返回對應的玩家，沒有則返回0，參見代碼如下：

 

解法一：

class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n) {
        board.resize(n, vector<int>(n, 0));   
    }

    int move(int row, int col, int player) {
        board[row][col] = player;
        int i = 0, j = 0, n = board.size();
        for (j = 1; j < n; ++j) {
            if (board[row][j] != board[row][j - 1]) break;
        }
        if (j == n) return player;
        for (i = 1; i < n; ++i) {
            if (board[i][col] != board[i - 1][col]) break;
        }
        if (i == n) return player;
        if (row == col) {
            for (i = 1; i < n; ++i) {
                if (board[i][i] != board[i - 1][i - 1]) break;
            }
            if (i == n) return player;
        }
        if (row + col == n - 1) {
            for (i = 1; i < n; ++i) {
                if (board[n - i - 1][i] != board[n - i][i - 1]) break;
            }
            if (i == n) return player;
        }
        return 0;
    }
    
private:
    vector<vector<int>> board;
};

 

Follow up中讓我們用更高效的方法，那么根據提示中的，我們建立一個大小為n的一維數組rows和cols，還有變量對角線diag和逆對角線rev_diag，這種方法的思路是，如果玩家1在第一行某一列放了一個子，那么rows[0]自增1，如果玩家2在第一行某一列放了一個子，則rows[0]自減1，那么只有當rows[0]等於n或者-n的時候，表示第一行的子都是一個玩家放的，則游戲結束返回該玩家即可，其他各行各列，對角線和逆對角線都是這種思路，參見代碼如下：

 

解法二：

class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n): rows(n), cols(n), N(n), diag(0), rev_diag(0) {}

    int move(int row, int col, int player) {
        int add = player == 1 ? 1 : -1;
        rows[row] += add; 
        cols[col] += add;
        diag += (row == col ? add : 0);
        rev_diag += (row == N - col - 1 ? add : 0);
        return (abs(rows[row]) == N || abs(cols[col]) == N || abs(diag) == N || abs(rev_diag) == N) ? player : 0;
    }

private:
    vector<int> rows, cols;
    int diag, rev_diag, N;
};

 

參考資料：

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

https://discuss.leetcode.com/topic/44548/java-o-1-solution-easy-to-understand

https://discuss.leetcode.com/topic/44605/c-time-o-1-space-o-n-short-simple-solution

 

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