[Leetcode] 794. Valid Tic-Tac-Toe State 解题报告

一、题目描述：

大致意思是：判定棋盘的状态是否符合规则

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"}

二、题目分析

解决此问题思路很明确，找出几种状态的判定条件，然后对输入进行判断。根据给出的案例。

（1）X，Y的输入顺序，即X最多比O多一个，或者相等。

Input: board = ["O  ", "   ", "   "]

这种情况O比X多所以输出false

（2）棋盘只能存在一个赢家

Input: board = ["XXX", "   ", "OOO"]
这种存在两个赢家也是False

（3）题目没有给出的一种具体情况

　　　

X X X
O O _
O _ _

当X赢得胜利时，应停止输入所以应该有两个。
三、技巧与编码
（1） 表示输入顺序用一个turn 来表示，若为X则加O则减判断最后 turn的值若为0或者1则为正常。
（2）用一次两层循环记录所有的状态，整合变量的思想，记录某一行某一列的状态而不是中情况的状态。
c++代码如下：
class Solution {
public:
    bool validTicTacToe(vector<string>& board) {
        //定义变量
        int turn=0;
        int row[3]={0},col[3]={0};
        int diag=0,antidiag=0;
        bool winx,wino;
        for(int i=0;i<=2;i++)
        {
            for(int j=0;j<=2;j++)
            {
                if(board[i][j]=='X')
                {
                    row[i]++;
                    col[j]++;
                    turn++;
                    if(i==j)
                    {
                        diag++;
                        
                    }
                    if(i+j==2)
                    {
                        antidiag++;
                    }
                    
                }
                else if(board[i][j]=='O')
                {
                    row[i]--;
                    col[j]--;
                    turn--;
                    if(i==j)
                    {
                        diag--;
                        
                    }
                    if(i+j==2)
                    {
                        antidiag--;
                    }
                    
                    
                    
                }
                
                
            }
        }
        winx = row[0] == 3 || row[1] == 3 || row[2] == 3 ||
               col[0] == 3 || col[1] == 3 || col[2] == 3 ||
               diag == 3 || antidiag == 3;
        
    wino = row[0]==-3 || row[1] == -3 || row[2] == -3 ||
               col[0] == -3 || col[1] == -3 || col[2] == -3 ||
               diag == -3 || antidiag == -3;
        if((winx&&turn==0) || (wino&&turn==1))
        {
            return false;
        }
        return (turn==0||turn==1)&&(!winx||!wino);//当顺序没有错误，并且两者都没有全部胜利的时候。
        
    }
};