[leetcode] 794. Valid Tic-Tac-Toe State

[leetcode] 794. Valid Tic-Tac-Toe State

Description

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”}.

分析

题目的意思是：验证#字棋的棋盘状态是否合法。

由规则可知，”X”一定最先开始，所以当前局面存在”O”的个数大于”X”的个数为非法。其次，由于”X”和”O”轮流，因此，当前局面中”X”的个数要么和”O”相等，要么比”O”多一。“O”在当前局面赢得比赛的情况下，上一轮的”X”一定不能赢得局面。“O”在当前局面赢得比赛的情况下，上一轮的”X”没有赢得局面时，合法局面必须满足”O”的个数等于”X”的个数。“X”在当前局面赢得比赛的情况下，意味着上一轮”O”没有赢得局面，合法局面下，”X”的个数正好比”O”的个数多一。

代码

class Solution {public: bool validTicTacToe(vector& board) { int cx=cnt(board,'X'), co=cnt(board,'O'); bool wx=win(board,'X'), wo=win(board,'O'); if(wx&&wo) return false; if(wx) return cx==co+1; if(wo) return cx==co; return cx==co+1||co==cx; } int cnt(vector b,char c){ int res=0; for(string s:b){ res+=count(s.begin(),s.end(),c); } return res; } bool win(vector b, char c){ for(int i=0;i<3;i++){ if(b[i][0]==c&&b[i][1]==c&&b[i][2]==c){ return true; } } for(int j=0;j<3;j++){ if(b[0][j]==c&&b[1][j]==c&&b[2][j]==c){ return true; } } return (b[0][0]==c&&b[1][1]==c&&b[2][2]==c)||(b[0][2]==c&&b[1][1]==c&&b[2][0]==c); }};

参考文献

​​794. Valid Tic-Tac-Toe State​​​​LWC 74: 794. Valid Tic-Tac-Toe State​​

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