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150 lines
5.0 KiB
Markdown
150 lines
5.0 KiB
Markdown
# [1275. Find Winner on a Tic Tac Toe Game](https://leetcode.com/problems/find-winner-on-a-tic-tac-toe-game/)
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## 题目
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Tic-tac-toe is played by two players *A* and *B* on a 3 x 3 grid.
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Here are the rules of Tic-Tac-Toe:
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- Players take turns placing characters into empty squares (" ").
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- The first player *A* always places "X" characters, while the second player *B* always places "O" characters.
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- "X" and "O" characters are always placed into empty squares, never on filled ones.
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- The game ends when there are 3 of the same (non-empty) character filling any row, column, or diagonal.
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- The game also ends if all squares are non-empty.
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- No more moves can be played if the game is over.
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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.
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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".
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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**.
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**Example 1**:
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```
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Input: moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
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Output: "A"
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Explanation: "A" wins, he always plays first.
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"X " "X " "X " "X " "X "
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" " -> " " -> " X " -> " X " -> " X "
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" " "O " "O " "OO " "OOX"
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```
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**Example 2**:
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```
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Input: moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
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Output: "B"
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Explanation: "B" wins.
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"X " "X " "XX " "XXO" "XXO" "XXO"
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" " -> " O " -> " O " -> " O " -> "XO " -> "XO "
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" " " " " " " " " " "O "
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```
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**Example 3**:
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```
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Input: moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
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Output: "Draw"
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Explanation: The game ends in a draw since there are no moves to make.
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"XXO"
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"OOX"
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"XOX"
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```
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**Example 4**:
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```
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Input: moves = [[0,0],[1,1]]
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Output: "Pending"
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Explanation: The game has not finished yet.
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"X "
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" O "
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" "
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```
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**Constraints:**
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- `1 <= moves.length <= 9`
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- `moves[i].length == 2`
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- `0 <= moves[i][j] <= 2`
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- There are no repeated elements on `moves`.
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- `moves` follow the rules of tic tac toe.
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## 题目大意
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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：
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- 玩家轮流将棋子放在空方格 (" ") 上。
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- 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。
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- "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。
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- 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。
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- 如果所有方块都放满棋子（不为空），游戏也会结束。
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- 游戏结束后，棋子无法再进行任何移动。
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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。
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提示：
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- 1 <= moves.length <= 9
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- moves[i].length == 2
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- 0 <= moves[i][j] <= 2
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- moves 里没有重复的元素。
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- moves 遵循井字棋的规则。
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## 解题思路
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- 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。
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- 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。
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## 代码
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```go
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func tictactoe(moves [][]int) string {
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board := [3][3]byte{}
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for i := 0; i < len(moves); i++ {
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if i%2 == 0 {
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board[moves[i][0]][moves[i][1]] = 'X'
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} else {
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board[moves[i][0]][moves[i][1]] = 'O'
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}
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}
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for i := 0; i < 3; i++ {
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if board[i][0] == 'X' && board[i][1] == 'X' && board[i][2] == 'X' {
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return "A"
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}
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if board[i][0] == 'O' && board[i][1] == 'O' && board[i][2] == 'O' {
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return "B"
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}
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if board[0][i] == 'X' && board[1][i] == 'X' && board[2][i] == 'X' {
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return "A"
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}
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if board[0][i] == 'O' && board[1][i] == 'O' && board[2][i] == 'O' {
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return "B"
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}
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}
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if board[0][0] == 'X' && board[1][1] == 'X' && board[2][2] == 'X' {
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return "A"
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}
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if board[0][0] == 'O' && board[1][1] == 'O' && board[2][2] == 'O' {
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return "B"
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}
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if board[0][2] == 'X' && board[1][1] == 'X' && board[2][0] == 'X' {
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return "A"
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}
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if board[0][2] == 'O' && board[1][1] == 'O' && board[2][0] == 'O' {
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return "B"
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}
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if len(moves) < 9 {
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return "Pending"
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}
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return "Draw"
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}
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``` |