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[Oracle] LeetCode 348 Design Tic-Tac-Toe

本文主要是介绍[Oracle] LeetCode 348 Design Tic-Tac-Toe，对大家解决编程问题具有一定的参考价值，需要的程序猿们随着小编来一起学习吧！

Assume the following rules are for the tic-tac-toe game on an n x n board between two players:

  • A move is guaranteed to be valid and is placed on an empty block.
  • Once a winning condition is reached, no more moves are allowed.
  • A player who succeeds in placing n of their marks in a horizontal, vertical, or diagonal row wins the game.

Implement the TicTacToe class:

  • TicTacToe(int n) Initializes the object the size of the board n.
  • int move(int row, int col, int player) Indicates that the player with id player plays at the cell (row, col) of the board. The move is guaranteed to be a valid move.

Solution

按照正常思路，对每行，每列，对角线都 \(check\) 的话，每次都得 \(O(N)\). 但实际上我们只关心出现的次数是否为 \(n\) 即可，所以为每个 \(player\) 分别开个 \(vector\) 记录即可

点击查看代码 
class TicTacToe {
private:
    vector<vector<int>> R, C;
    vector<int> diag, r_diag;
    
public:
    TicTacToe(int n) {
        R = vector<vector<int>> (n, vector<int>(3));
        C = vector<vector<int>> (n, vector<int>(3));
        diag = vector<int> (3);
        r_diag = vector<int> (3);
    }
    
    int move(int row, int col, int player) {
        int cnt;
        R[row][player]++; C[col][player]++;
        cnt = max(R[row][player], C[col][player]);
        if(row==col)diag[player]++;
        if(row+col==R.size()-1)r_diag[player]++;
        cnt=max(cnt, max(diag[player], r_diag[player]));
        if(cnt==R.size())return player;
        else return 0;
    }
};

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