1275. Find Winner on a Tic Tac Toe Game

# 1275. Find Winner on a Tic Tac Toe Game#

## 题目 #

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:

• 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 3 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 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 <= 9`
• `moves[i].length == 2`
• `0 <= moves[i][j] <= 2`
• There are no repeated elements on `moves`.
• `moves` follow the rules of tic tac toe.

## 题目大意 #

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：

• 玩家轮流将棋子放在空方格 (” “) 上。
• 第一个玩家 A 总是用 “X” 作为棋子，而第二个玩家 B 总是用 “O” 作为棋子。
• “X” 和 “O” 只能放在空方格中，而不能放在已经被占用的方格上。
• 只要有 3 个相同的（非空）棋子排成一条直线（行、列、对角线）时，游戏结束。
• 如果所有方块都放满棋子（不为空），游戏也会结束。
• 游戏结束后，棋子无法再进行任何移动。

• 1 <= moves.length <= 9
• moves[i].length == 2
• 0 <= moves[i][j] <= 2
• moves 里没有重复的元素。
• moves 遵循井字棋的规则。

## 解题思路 #

• 两人玩 3*3 井字棋，A 先走，B 再走。谁能获胜就输出谁，如果平局输出 “Draw”，如果游戏还未结束，输出 “Pending”。游戏规则：谁能先占满行、列或者对角线任意一条线，谁就赢。
• 简单题。题目给定 move 数组最多 3 步，而要赢得比赛，必须走满 3 步，所以可以先模拟，按照给的步数数组把 A 和 B 的步数都放在棋盘上。然后依次判断行、列，对角线的三种情况。如果都判完了，剩下的情况就是平局和死局的情况。

## 代码 #

``````
package leetcode

func tictactoe(moves [][]int) string {
board := byte{}
for i := 0; i < len(moves); i++ {
if i%2 == 0 {
board[moves[i]][moves[i]] = 'X'
} else {
board[moves[i]][moves[i]] = 'O'
}
}
for i := 0; i < 3; i++ {
if board[i] == 'X' && board[i] == 'X' && board[i] == 'X' {
return "A"
}
if board[i] == 'O' && board[i] == 'O' && board[i] == 'O' {
return "B"
}
if board[i] == 'X' && board[i] == 'X' && board[i] == 'X' {
return "A"
}
if board[i] == 'O' && board[i] == 'O' && board[i] == 'O' {
return "B"
}
}
if board == 'X' && board == 'X' && board == 'X' {
return "A"
}
if board == 'O' && board == 'O' && board == 'O' {
return "B"
}
if board == 'X' && board == 'X' && board == 'X' {
return "A"
}
if board == 'O' && board == 'O' && board == 'O' {
return "B"
}
if len(moves) < 9 {
return "Pending"
}
return "Draw"
}

`````` Apr 8, 2023 Edit this page