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119 lines
4.0 KiB
Markdown
119 lines
4.0 KiB
Markdown
# [794. Valid Tic-Tac-Toe State](https://leetcode.com/problems/valid-tic-tac-toe-state/)
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## 题目
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Given a Tic-Tac-Toe board as a string array board, return true if and only if it is possible to reach this board position during the course of a valid tic-tac-toe game.
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The board is a 3 x 3 array that consists of characters ' ', 'X', and 'O'. The ' ' character represents an empty square.
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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 always places 'X' characters, while the second player always places 'O' characters.
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- 'X' and 'O' characters are always placed into empty squares, never filled ones.
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- The game ends when there are three 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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**Example 1**:
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Input: board = ["O "," "," "]
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Output: false
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Explanation: The first player always plays "X".
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**Example 2**:
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Input: board = ["XOX"," X "," "]
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Output: false
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Explanation: Players take turns making moves.
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**Example 3**:
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Input: board = ["XXX"," ","OOO"]
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Output: false
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**Example 4**:
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Input: board = ["XOX","O O","XOX"]
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Output: true
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**Constraints:**
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- board.length == 3
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- board[i].length == 3
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- board[i][j] is either 'X', 'O', or ' '.
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## 题目大意
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给你一个字符串数组 board 表示井字游戏的棋盘。当且仅当在井字游戏过程中，棋盘有可能达到 board 所显示的状态时，才返回 true 。
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井字游戏的棋盘是一个 3 x 3 数组，由字符 ' '，'X' 和 'O' 组成。字符 ' ' 代表一个空位。
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以下是井字游戏的规则：
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- 玩家轮流将字符放入空位（' '）中。
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- 玩家 1 总是放字符 'X' ，而玩家 2 总是放字符 'O' 。
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- 'X' 和 'O' 只允许放置在空位中，不允许对已放有字符的位置进行填充。
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- 当有 3 个相同（且非空）的字符填充任何行、列或对角线时，游戏结束。
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- 当所有位置非空时，也算为游戏结束。
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- 如果游戏结束，玩家不允许再放置字符。
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## 解题思路
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分类模拟：
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- 根据题意棋盘在任意时候，要么 X 的数量比 O 的数量多 1，要么两者相等
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- X 的数量等于 O 的数量时,任何行、列或对角线都不会出现 3 个相同的 X
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- X 的数量比 O 的数量多 1 时,任何行、列或对角线都不会出现 3 个相同的 O
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## 代码
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```go
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package leetcode
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func validTicTacToe(board []string) bool {
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cntX, cntO := 0, 0
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for i := range board {
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for j := range board[i] {
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if board[i][j] == 'X' {
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cntX++
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} else if board[i][j] == 'O' {
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cntO++
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}
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}
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}
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if cntX < cntO || cntX > cntO+1 {
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return false
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}
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if cntX == cntO {
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return process(board, 'X')
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}
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return process(board, 'O')
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}
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func process(board []string, c byte) bool {
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//某一行是"ccc"
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if board[0] == string([]byte{c, c, c}) || board[1] == string([]byte{c, c, c}) || board[2] == string([]byte{c, c, c}) {
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return false
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}
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//某一列是"ccc"
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if (board[0][0] == c && board[1][0] == c && board[2][0] == c) ||
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(board[0][1] == c && board[1][1] == c && board[2][1] == c) ||
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(board[0][2] == c && board[1][2] == c && board[2][2] == c) {
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return false
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}
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//某一对角线是"ccc"
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if (board[0][0] == c && board[1][1] == c && board[2][2] == c) ||
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(board[0][2] == c && board[1][1] == c && board[2][0] == c) {
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return false
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}
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return true
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}
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``` |