794. Valid Tic-Tac-Toe State

 

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"}.

 

这道题又是关于井字棋游戏的,之前也有一道类似的题 Design Tic-Tac-Toe,不过那道题是模拟游戏进行的,而这道题是让验证当前井字棋的游戏状态是否正确。这题的例子给的比较好,cover 了很多种情况:

 

情况一:

0 _ _
_ _ _
_ _ _

 

这是不正确的状态,因为先走的使用X,所以只出现一个O,是不对的。

 

情况二:

X O X
_ X _
_ _ _

 

这个也是不正确的,因为两个 player 交替下棋,X最多只能比O多一个,这里多了两个,肯定是不对的。

 

情况三:

X X X
_ _ _ 
O O O

 

这个也是不正确的,因为一旦第一个玩家的X连成了三个,那么游戏马上结束了,不会有另外一个O出现。

 

情况四:

X O X
O _ O
X O X

 

这个状态没什么问题,是可以出现的状态。

好,那么根据给的这些例子,可以分析一下规律,根据例子1和例子2得出下棋顺序是有规律的,必须是先X后O,不能破坏这个顺序,那么可以使用一个 turns 变量,当是X时,turns 自增1,反之若是O,则 turns 自减1,那么最终 turns 一定是0或者1,其他任何值都是错误的,比如例子1中,turns 就是 -1,例子2中,turns 是2,都是不对的。根据例子3,可以得出结论,只能有一个玩家获胜,可以用两个变量 xwin 和 owin,来记录两个玩家的获胜状态,由于井字棋的制胜规则是横竖斜任意一个方向有三个连续的就算赢,那么分别在各个方向查找3个连续的X,有的话 xwin 赋值为 true,还要查找3个连续的O,有的话 owin 赋值为 true,例子3中 xwin 和 owin 同时为 true 了,是错误的。还有一种情况,例子中没有 cover 到的是:

 

情况五:

X X X
O O _
O _ _

 

这里虽然只有 xwin 为 true,但是这种状态还是错误的,因为一旦第三个X放下后,游戏立即结束,不会有第三个O放下,这么检验这种情况呢?这时 turns 变量就非常的重要了,当第三个O放下后,turns 自减1,此时 turns 为0了,而正确的应该是当 xwin 为 true 的时候,第三个O不能放下,那么 turns 不减1,则还是1,这样就可以区分情况五了。当然,可以交换X和O的位置,即当 owin 为 true 时,turns 一定要为0。现在已经覆盖了搜索的情况了,参见代码如下:

 

class Solution {
public:
    bool validTicTacToe(vector<string>& board) {
        bool xwin = false, owin = false;
        vector<int> row(3), col(3);
        int diag = 0, antidiag = 0, turns = 0;
        for (int i = 0; i < 3; ++i) {
            for (int j = 0; j < 3; ++j) {
                if (board[i][j] == 'X') {
                    ++row[i]; ++col[j]; ++turns;
                    if (i == j) ++diag;
                    if (i + j == 2) ++antidiag;
                } else if (board[i][j] == 'O') {
                    --row[i]; --col[j]; --turns;
                    if (i == j) --diag;
                    if (i + j == 2) --antidiag;
                }
            }
        }
        xwin = row[0] == 3 || row[1] == 3 || row[2] == 3 ||
               col[0] == 3 || col[1] == 3 || col[2] == 3 ||
               diag == 3 || antidiag == 3;
        owin = row[0] == -3 || row[1] == -3 || row[2] == -3 ||
               col[0] == -3 || col[1] == -3 || col[2] == -3 ||
               diag == -3 || antidiag == -3;
        if ((xwin && turns == 0) || (owin && turns == 1)) return false;
        return (turns == 0 || turns == 1) && (!xwin || !owin);
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/794

 

类似题目:

Design Tic-Tac-Toe

 

参考资料:

https://leetcode.com/problems/valid-tic-tac-toe-state/

https://leetcode.com/problems/valid-tic-tac-toe-state/discuss/117580/Straightforward-Java-solution-with-explaination

 

LeetCode All in One 题目讲解汇总(持续更新中…)


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