leetcode/1275.Find-Winner-on-a-Tic-Tac-Toe-Game/README.md
Tic-tac-toe is played by two players A and B on a 3 x 3 grid.
Here are the rules of Tic-Tac-Toe:
Given an array moves where each element is another array of size 2 corresponding to the row and column of the grid where they mark their respective character in the order in which A and B play.
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 (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, he always plays first.
"X " "X " "X " "X " "X "
" " -> " " -> " X " -> " X " -> " X "
" " "O " "O " "OO " "OOX"
Example 2:
Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
Output: "B"
Explanation: "B" wins.
"X " "X " "XX " "XXO" "XXO" "XXO"
" " -> " O " -> " O " -> " O " -> "XO " -> "XO "
" " " " " " " " " " "O "
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.
"XXO"
"OOX"
"XOX"
Example 4:
Input: moves = [[0,0],[1,1]]
Output: "Pending"
Explanation: The game has not finished yet.
"X "
" O "
" "
Constraints:
1 <= moves.length <= 9moves[i].length == 20 <= moves[i][j] <= 2moves.moves follow the rules of tic tac toe.A 和 B 在一个 3 x 3 的网格上玩井字棋。井字棋游戏的规则如下:
给你一个数组 moves,其中每个元素是大小为 2 的另一个数组(元素分别对应网格的行和列),它按照 A 和 B 的行动顺序(先 A 后 B)记录了两人各自的棋子位置。如果游戏存在获胜者(A 或 B),就返回该游戏的获胜者;如果游戏以平局结束,则返回 "Draw";如果仍会有行动(游戏未结束),则返回 "Pending"。你可以假设 moves 都 有效(遵循井字棋规则),网格最初是空的,A 将先行动。
提示:
func tictactoe(moves [][]int) string {
board := [3][3]byte{}
for i := 0; i < len(moves); i++ {
if i%2 == 0 {
board[moves[i][0]][moves[i][1]] = 'X'
} else {
board[moves[i][0]][moves[i][1]] = 'O'
}
}
for i := 0; i < 3; i++ {
if board[i][0] == 'X' && board[i][1] == 'X' && board[i][2] == 'X' {
return "A"
}
if board[i][0] == 'O' && board[i][1] == 'O' && board[i][2] == 'O' {
return "B"
}
if board[0][i] == 'X' && board[1][i] == 'X' && board[2][i] == 'X' {
return "A"
}
if board[0][i] == 'O' && board[1][i] == 'O' && board[2][i] == 'O' {
return "B"
}
}
if board[0][0] == 'X' && board[1][1] == 'X' && board[2][2] == 'X' {
return "A"
}
if board[0][0] == 'O' && board[1][1] == 'O' && board[2][2] == 'O' {
return "B"
}
if board[0][2] == 'X' && board[1][1] == 'X' && board[2][0] == 'X' {
return "A"
}
if board[0][2] == 'O' && board[1][1] == 'O' && board[2][0] == 'O' {
return "B"
}
if len(moves) < 9 {
return "Pending"
}
return "Draw"
}