LeetCode1275. Find Winner on a Tic Tac Toe Game

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一、题目

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.

二、题解

class Solution {
public:string tictactoe(vector<vector<int>>& moves) {int n = moves.size();vector<vector<string>> grid(3,vector<string>(3," "));for(int i = 0;i < n;i++){if(i % 2 == 0) grid[moves[i][0]][moves[i][1]] = "X";else grid[moves[i][0]][moves[i][1]] = "O";}//判断行for(int i = 0;i < 3;i++){if(grid[i][0] != " " && grid[i][0] == grid[i][1] && grid[i][1] == grid[i][2]){if(grid[i][0] == "X") return "A";else return "B";}}//判断列for(int j = 0;j < 3;j++){if(grid[0][j] != " " && grid[0][j] == grid[1][j] && grid[1][j] == grid[2][j]){if(grid[0][j] == "X") return "A";else return "B";}}//判断主对角线if(grid[0][0] != " " && grid[0][0] == grid[1][1] && grid[1][1] == grid[2][2]){if(grid[0][0] == "X") return "A";else return "B";}//判断副对角线if(grid[0][2] != " " && grid[0][2] == grid[1][1] && grid[1][1] == grid[2][0]){if(grid[0][2] == "X") return "A";else return "B";}if(moves.size() < 9) return "Pending";else return "Draw";}
};

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