Tic-tac-toeis played by two playersAandBon a3 x 3grid. The rules of Tic-Tac-Toe are:

• Players take turns placing characters into empty squares' '.
• The first playerAalways places'X'characters, while the second playerBalways places'O'characters.
• 'X'and'O'characters are always placed into empty squares, never on filled ones.
• The game ends when there arethreeof 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.

Given a 2D integer arraymoveswheremoves[i] = [rowi, coli]indicates that theithmove will be played ongrid[rowi][coli]

You can assume thatmovesis valid (i.e., it follows the rules ofTic-Tac-Toe), the grid is initially empty, andAwill play first.

Example 1:

Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
Output: "A"
Explanation: A wins, they always play first.
 

Example 2:

Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: B wins.

Example 3:

Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
Output: "Draw"
Explanation: The game ends in a draw since there are no moves to make.

Example 4:

Input: moves = [[0,0],[1,1]]
Output: "Pending"
Explanation: The game has not finished yet.

Constraints:

• 1 <= moves.length <= 9
• moves[i].length == 2
• 0 <= rowi, coli<= 2
• There are no repeated elements onmoves.
• movesfollow the rules of tic tac toe.

這道題是關於井字棋遊戲的，這個遊戲簡單容易上手，是課間十五分鐘必備遊戲之一，遊戲規則很簡單，一個人畫叉，一個人畫圓圈，只要有三個相連的位置，包括對角線就算贏。而這道題給了兩個玩家的若干操作，讓判斷當前的棋盤狀態，是哪一方贏了，還是沒下完，或者是平局。判贏的條件就是找到任意行或列，或者對角線有三個相同的符號，若找不到有可能是平局或者沒下完，只要判斷總步數是否為9，就知道有沒有下完了。由於給定的是兩個玩家按順序落子的位置，一個比較直接的方法就是分別還原出兩個玩家在棋盤上的落子，分別還原出兩個 3 by 3 的棋盤的好處是可以不用區分到底是叉還是圓圈，只要找三個連續的落子位置就行了，而且可以把查詢邏輯放到一個子函式中進行復用。在子函式中，判斷各行各列是否有連續三個落子，以及兩條對角線，若有的話返回 true，否則 false。然後分別對還原出來的陣列A和B呼叫子函式，若有返回的 true，則返回對應的A或B。否則就判斷 moves 陣列的長度，若等於9，返回平局 Draw，反之為 Pending，參見程式碼如下：

解法一：

class Solution {
public:
    string tictactoe(vector<vector<int>>& moves) {
        vector<vector<int>> A(3, vector<int>(3)), B = A;
        for (int i = 0; i < moves.size(); ++i) {
            if (i % 2 == 0) A[moves[i][0]][moves[i][1]] = 1;
            else B[moves[i][0]][moves[i][1]] = 1;
        }
        if (helper(A)) return "A";
        if (helper(B)) return "B";
        return (moves.size() == 9) ? "Draw" : "Pending";
    }
    bool helper(vector<vector<int>>& v) {
        for (int i = 0; i < 3; ++i) {
            if (v[i][0] == 1 && v[i][1] == 1 && v[i][2] == 1) return true;
            if (v[0][i] == 1 && v[1][i] == 1 && v[2][i] == 1) return true;
        }
        if (v[0][0] == 1 && v[1][1] == 1 && v[2][2] == 1) return true;
        if (v[2][0] == 1 && v[1][1] == 1 && v[0][2] == 1) return true;
        return false;
    }
};

再來看一種更簡潔的寫法，為了更好的區分玩家A和B落子，這裡可以用1和 -1 來表示，這樣若玩家A有三個相連的落子，則和為3，而玩家B的相連3個落子之和為 -3，只要判斷各行各列以及對角線的和的絕對值，若為3，則肯定有玩家獲勝，具體區分是哪個玩家，可以根據當前運算元的位置來判斷，遍歷 moves 陣列時，若下標為偶數，則表示是玩家A，奇數則為玩家B，根據i的奇偶行來賦值 val 為1或 -1。取出落子位置的橫縱座標r和c，若二者相等，則是主對角線，diag 加上 val，若二者和為2，則是逆對角線，rev_diag 加上 val。然後 row 和 col 陣列對應的位置加上 val，接下來判斷行列對角線的和的絕對值是否有等於3的，有的話就根據 val 的正負來返回玩家A或B。否則就判斷 moves 陣列的長度，若等於9，返回平局 Draw，反之為 Pending，參見程式碼如下：

解法二：

class Solution {
public:
    string tictactoe(vector<vector<int>>& moves) {
        int diag = 0, rev_diag = 0;
        vector<int> row(3), col(3);
        for (int i = 0; i < moves.size(); ++i) {
            int r = moves[i][0], c = moves[i][1], val = i % 2 == 0 ? 1 : -1;
            if (r == c) diag += val;
            if (r + c == 2) rev_diag += val;
            row[r] += val;
            col[c] += val;
            if (abs(diag) == 3 || abs(rev_diag) == 3 || abs(row[r]) == 3 || abs(col[c]) == 3) return val == 1 ? "A" : "B";
        }
        return moves.size() == 9 ? "Draw" : "Pending";
    }
};

Github 同步地址:

https://github.com/grandyang/leetcode/issues/1275

參考資料：

https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/

https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/discuss/638224/Short-Java-code-using-diagonals-and-colrow-arrays

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