LeetCode1275. Find Winner on a Tic Tac Toe Game

文章目录

- 一、题目
- 二、题解

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

cpp 复制代码

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";
    }
};