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：
_ _
_ _ _
_ _ _
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(), col();
int diag = , antidiag = , turns = ;
for (int i = ; i < ; ++i) {
for (int j = ; j < ; ++j) {
if (board[i][j] == 'X') {
++row[i]; ++col[j]; ++turns;
if (i == j) ++diag;
if (i + j == ) ++antidiag;
} else if (board[i][j] == 'O') {
--row[i]; --col[j]; --turns;
if (i == j) --diag;
if (i + j == ) --antidiag;
}
}
}
xwin = row[] == || row[] == || row[] == ||
col[] == || col[] == || col[] == ||
diag == || antidiag == ;
owin = row[] == - || row[] == - || row[] == - ||
col[] == - || col[] == - || col[] == - ||
diag == - || antidiag == -;
if ((xwin && turns == ) || (owin && turns == )) return false;
return (turns == || turns == ) && (!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
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