leetcode 348. Design Tic-Tac-Toe【模拟】

348. Design Tic-Tac-Toe

Medium

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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:

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?

Accepted

54,270

Submissions

106,969

n*n的棋盘 两个人轮流玩

是否可以连着线 算赢

时间复杂度小一些

就想到分成两个数组存储了

class TicTacToe {
public:
    /** Initialize your data structure here. */
    vectorhor[2];
    vectorver[2];
    int dia[2][2];
    int size;
    TicTacToe(int n) {
        for(int i = 0; i < n; i++) {
            hor[0].push_back(0);
            hor[1].push_back(0);
            ver[0].push_back(0);
            ver[1].push_back(0);
        }
        size = n;
        dia[0][0] = 0;
        dia[1][0] = 0;
        dia[0][1] = 0;
        dia[1][1] = 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. */
    bool judge (int role) {
        for (int i = 0; i < size; i ++) {
            //printf("i=%d,hor=%d,ver=%d\n",i,hor[role-1][i],ver[role-1][i]);
            if (hor[role - 1][i] == size || ver[role - 1][i] == size) {
                return true;
            }
        }
        if (dia[role -1][0] == size || dia[role-1][1] == size)
            return true;
        return false;
    }
    int move(int row, int col, int player) {
        hor[player - 1][row] ++;
        ver[player - 1][col] ++;
        if (row == col) {
            dia[player - 1][0] ++;
        }
        if (row + col == size - 1) {
            dia[player - 1][1] ++;
        }
        if (judge(player)) 
            return player;
        return false;
    }
};

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