Programmer Interview Gold Code-Interview Question 16.04. Tic-Tac-Toe (Counting)

1. Title

Design an algorithm to determine whether the player has won the tic-tac-toe game. The input is an N x N array checkerboard, consisting of the characters "", "X" and "O", where the character "" represents an empty slot.

Here are the rules of the tic-tac-toe game:

• Players take turns putting characters into empty slots ("").
• The first player always puts the character "O", and the second player always puts the character "X".
• "X" and "O" are only allowed to be placed in vacancies, and not allowed to fill the positions where characters have been placed.
• When there are N identical (and non-empty) characters filling any row, column, or diagonal, the game ends and the player corresponding to the character wins.
• When all positions are not empty, it is also regarded as the end of the game.
• If the game is over, the player is not allowed to place any more characters.

If there is a winner in the game, the character used by the winner of the game ("X" or "O")
is returned ; if the game ends in a draw, then "Draw" is returned;
if there is still action (the game is not over), then Return to "Pending".

示例 1：
输入： board = ["O X"," XO","X O"]
输出： "X"

示例 2：
输入： board = ["OOX","XXO","OXO"]
输出： "Draw"
解释： 没有玩家获胜且不存在空位

示例 3：
输入： board = ["OOX","XXO","OX "]
输出： "Pending"
解释： 没有玩家获胜且仍存在空位

提示：
1 <= board.length == board[i].length <= 100
输入一定遵循井字棋规则

Source: LeetCode
Link: https://leetcode-cn.com/problems/tic-tac-toe-lcci
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2. Problem solving

• Similar topics: LeetCode 1275. Find the winner of Tic Tac Toe (bit operation)
• Count by row, column, diagonal

class Solution {
public:
    string tictactoe(vector<string>& board) {
        int n = board.size(), i, j, count = 0;
        vector<int> ra(n,0);//玩家1 的行
        vector<int> ca(n,0);//玩家1 的列
        vector<int> rb(n,0);
        vector<int> cb(n,0);
        int xa1 = 0, xa2 = 0, xb1 = 0, xb2 = 0;//对角线2条
        for(i = 0; i < n; ++i)
        {
            for(j = 0; j < n; ++j)
            {
                if(board[i][j]=='X')
                {
                    ra[i]++;
                    ca[j]++;
                    if(i==j) xa1++;
                    if(i+j==n-1) xa2++;
                    count++;
                }
                else if(board[i][j]=='O')
                {
                    rb[i]++;
                    cb[j]++;
                    if(i==j) xb1++;
                    if(i+j==n-1) xb2++;
                    count++;
                }
            }
        }
        char win = '-';
        for(i = 0; i < n; ++i)
        {
            if(ra[i]==n || ca[i]==n ||xa1==n || xa2==n)
            {
                win = 'X';
                break;
            }
            if(rb[i]==n || cb[i]==n || xb1==n || xb2==n)
            {
                win = 'O';
                break;
            }
        }
        if(win=='X') return "X";
        if(win=='O') return "O";
        if(count == n*n) return "Draw";
        return "Pending";
    }
};

8 ms	8.3 MB