[Leetcode] 348. Design Tic-Tac-Toe 解题报告

题目：

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() operation?

思路：

这道题目的思路没有什么特别奇怪的。只是我开始用了二维数组来表示棋盘，这样空间复杂度就成为O(n^2)，move函数的时间复杂度是O(n)。后来发现一种更高效的实现方法：用两个数组和两个整数表示一个player目前达到的状态。这样空间复杂度就可以降低到O(n)，而move的时间复杂度竟然可以降低到O(1)。

这种面试题到处都是坑啊！最优算法往往未必就能一下子想到。

代码：

class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n) {
        rows_first.resize(n, 0);
        rows_second.resize(n, 0);
        cols_first.resize(n, 0);
        cols_second.resize(n, 0);
        diag_first = diag_second = anti_diag_first = anti_diag_second = 0;
        size = 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. */
    int move(int row, int col, int player) {
        if(player == 1) {
            if(++rows_first[row] == size)                           return 1;
            if(++cols_first[col] == size)                           return 1;
            if(row == col && ++diag_first == size)                  return 1;
            if(row + col == size - 1 && ++anti_diag_first == size)  return 1;
        }
        else {
            if(++rows_second[row] == size)                          return 2;
            if(++cols_second[col] == size)                          return 2;
            if(row == col && ++diag_second == size)                 return 2;
            if(row + col == size - 1 && ++anti_diag_second == size) return 2;
        }
        return 0;
    }
private:
    vector rows_first, rows_second;
    vector cols_first, cols_second;
    int diag_first, diag_second, anti_diag_first, anti_diag_second;
    int size;
};

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