面试题 16.04. 井字游戏

用循环 暴力即可

class Solution {
public:
    string tictactoe(vector<string>& board) {
        int n = board.size();
        int flag1 = 0, flag2 = 0, flag3 = 1;
        for(int i = 0; i < n; i++)
        {
            char y = board[i][0];
            if(y == ' ')
            {
                flag3 = 0;
                continue;
            }
            int j = 1;
            for(; j < n; j++)
            {
                if(board[i][j] == ' ')
                {
                    flag3 = 0;
                    break;
                }
                if(board[i][j] != y)
                    break;
            }
            if(j == n)
            {
                if(y == 'X')
                    flag1 = 1;
                else flag2 = 1;
            }
        }
        for(int j = 0; j < n; j++)
        {
            char y = board[0][j];
            if(y == ' ')
            {
                flag3 = 0;
                continue;
            }
            int i = 1;
            for(; i < n; i++)
            {
                if(board[i][j] == ' ')
                {
                    flag3 = 0;
                    break;
                }
                if(board[i][j] != y)
                    break;
            }
            if(i == n)
            {
                if(y == 'X')
                    flag1 = 1;
                else flag2 = 1;
            }
        }
        char y = board[0][0];
        if(y == ' ')
            flag3 = 0;
        if(y != ' ')
        {
            int i = 1;
            for(; i < n; i++)
            {
                if(board[i][i] == ' ')
                {
                    flag3 = 0;
                    break;
                }
                if(board[i][i] != y)
                    break;
            }
            if(i == n)
            {
                if(y == 'X') flag1 = 1;
                else flag2 = 1;
            }
        }
        y = board[0][n - 1];
        if(y == ' ')
            flag3 = 0;
        if(y != ' ')
        {
            int i = 1;
            for(; i < n; i++)
            {
                if(board[i][n - i - 1] == ' ')
                {
                    flag3 = 0;
                    break;
                }
                if(board[i][n - i - 1] != y)
                    break;
            }
            if(i == n)
            {
                if(y == 'X') flag1 = 1;
                else flag2 = 1;
            }
        }

        if(flag1 == 1) return "X";
        if(flag2 == 1) return "O";
        return flag3 == 1 ? "Draw" : "Pending";
    }
};