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

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

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: 搞不懂为啥他说在O(n*n)以内解决, 感觉不可能到O(n*n)的复杂度啊, 每一步先判断横竖两行, 如果能够排满就赢了, 否则如果这个位置在两个对角线上再判断对角线是否排满即可. 时间复杂度O(n) or O(1).

代码如下:

class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n):mp(vector>(n, vector(n, 0))), 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. */
    int move(int row, int col, int player) {
        mp[row][col] = player;
        int i;
        for(i=0; i> mp;
    int N;
};

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

class TicTacToe {
public:
    /** Initialize your data structure here. */
    TicTacToe(int n):rows(vector(n, 0)), cols(vector(n, 0)) {
        
    }
    
    /** 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) {
        int flag = (player==1?1:-1), n = rows.size();
        rows[row] += flag, cols[col]+=flag;
        if(row==col) antiDiagnal += flag;
        if(row+col+1 == n) diagnal += flag;
        if(abs(rows[row])==n || abs(cols[col])==n 
          ||abs(diagnal)==n || abs(antiDiagnal)==n) 
            return player;
        return 0;
    }
private:
    vector rows, cols;
    int diagnal=0, antiDiagnal=0;
};

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