README_EN.md

1275. Find Winner on a Tic Tac Toe Game

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Description

Tic-tac-toe is played by two players A and B on a 3 x 3 grid. The rules of Tic-Tac-Toe are:

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

Given a 2D integer array moves where moves[i] = [rowi, coli] indicates that the ith move will be played on grid[rowi][coli]. return the winner of the game if it exists (A or B). In case the game ends in a draw return "Draw". If there are still movements to play return "Pending".

You can assume that moves is valid (i.e., it follows the rules of Tic-Tac-Toe), the grid is initially empty, and A will 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.

 

Constraints:

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

Solutions

Python3

class Solution:
    def tictactoe(self, moves: List[List[int]]) -> str:
        n = len(moves)
        counter = [0] * 8
        for i in range(n - 1, -1, -2):
            row, col = moves[i][0], moves[i][1]
            counter[row] += 1
            counter[col + 3] += 1
            if row == col:
                counter[6] += 1
            if row + col == 2:
                counter[7] += 1
            if (
                counter[row] == 3
                or counter[col + 3] == 3
                or counter[6] == 3
                or counter[7] == 3
            ):
                return "A" if (i % 2) == 0 else "B"
        return "Draw" if n == 9 else "Pending"

Java

class Solution {
    public String tictactoe(int[][] moves) {
        int n = moves.length;
        int[] counter = new int[8];
        for (int i = n - 1; i >= 0; i -= 2) {
            int row = moves[i][0], col = moves[i][1];
            ++counter[row];
            ++counter[col + 3];
            if (row == col) ++counter[6];
            if (row + col == 2) ++counter[7];
            if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
                return (i % 2) == 0 ? "A" : "B";
            }
        }
        return n == 9 ? "Draw" : "Pending";
    }
}

C++

class Solution {
public:
    string tictactoe(vector<vector<int>>& moves) {
        int n = moves.size();
        vector<int> counter(8, 0);
        for (int i = n - 1; i >= 0; i -= 2) {
            int row = moves[i][0], col = moves[i][1];
            ++counter[row];
            ++counter[col + 3];
            if (row == col) ++counter[6];
            if (row + col == 2) ++counter[7];
            if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
                return (i % 2 == 0) ? "A" : "B";
            }
        }
        return n == 9 ? "Draw" : "Pending";
    }
};

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