LeetCode #794 - Valid Tic-Tac-Toe State

最新推荐文章于 2021-12-09 21:36:53 发布

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本文介绍了一个算法，用于判断一个给定的3x3井字游戏（Tic-Tac-Toe）局面是否可以通过合法的游戏过程达到。通过计算X和O的数量，以及检查是否有胜利者，该算法能有效识别非法游戏状态。

题目描述：

A Tic-Tac-Toe board is given as a string array board. Return True if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.

The board is a 3 x 3 array, and consists of characters " ", "X", and "O". The " " character represents an empty square.

Here are the rules of Tic-Tac-Toe:

Players take turns placing characters into empty squares (" ").
The first player always places "X" characters, while the second player always places "O" characters.
"X" and "O" characters are always placed into empty squares, never filled ones.
The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
The game also ends if all squares are non-empty.
No more moves can be played if the game is over.

Example 1:
Input: board = ["O  ", "   ", "   "]
Output: false
Explanation: The first player always plays "X".

Example 2:
Input: board = ["XOX", " X ", "   "]
Output: false
Explanation: Players take turns making moves.

Example 3:
Input: board = ["XXX", "   ", "OOO"]
Output: false

Example 4:
Input: board = ["XOX", "O O", "XOX"]
Output: true

Note:

board is a length-3 array of strings, where each string board[i] has length 3.
Each board[i][j] is a character in the set {" ", "X", "O"}.

class Solution {
public:
    bool validTicTacToe(vector<string>& board) {
        int x_count=0;
        int o_count=0;
        for(int i=0;i<3;i++)
        {
            for(int j=0;j<3;j++)
            {
                if(board[i][j]=='X') x_count++;
                else if(board[i][j]=='O') o_count++;
            }
        }
        bool x_win=isWin(board,'X');
        bool o_win=isWin(board,'O');
        if(x_count<o_count||x_count>o_count+1) return false;
        else if(x_count==o_count&&x_win) return false; // 当X和O数目相同时，最后一步必定是O，那么在这之前棋局不能结束
        else if(x_count==o_count+1&&o_win) return false; // 当X比O数目大一时，最后一步必定是X，那么在这之前棋局不能结束
        return true;
    }
    
    bool isWin(vector<string>& board, char c)
    {
        for(int i=0;i<3;i++)
            if(board[i][0]==c&&board[i][0]==board[i][1]&&board[i][1]==board[i][2]) return true;    
        for(int j=0;j<3;j++)
            if(board[0][j]==c&&board[0][j]==board[1][j]&&board[1][j]==board[2][j]) return true;    
        if(board[0][0]==c&&board[0][0]==board[1][1]&&board[1][1]==board[2][2]) return true;  
        if(board[2][0]==c&&board[2][0]==board[1][1]&&board[1][1]==board[0][2]) return true; 
        return false;
    }
};