一刷794

Author: diguage

README.adoc

@@ -5583,14 +5583,14 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|{doc_base_url}/0793-preimage-size-of-factorial-zeroes-function.adoc[题解]
 //|Hard
 //|
-//
-//|{counter:codes}
-//|{leetcode_base_url}/valid-tic-tac-toe-state/[794. Valid Tic-Tac-Toe State^]
-//|{source_base_url}/_0794_ValidTicTacToeState.java[Java]
-//|{doc_base_url}/0794-valid-tic-tac-toe-state.adoc[题解]
-//|Medium
-//|
-//
+
+|{counter:codes}
+|{leetcode_base_url}/valid-tic-tac-toe-state/[794. Valid Tic-Tac-Toe State^]
+|{source_base_url}/_0794_ValidTicTacToeState.java[Java]
+|{doc_base_url}/0794-valid-tic-tac-toe-state.adoc[题解]
+|Medium
+|
+
 //|{counter:codes}
 //|{leetcode_base_url}/number-of-subarrays-with-bounded-maximum/[795. Number of Subarrays with Bounded Maximum^]
 //|{source_base_url}/_0795_NumberOfSubarraysWithBoundedMaximum.java[Java]

docs/0794-valid-tic-tac-toe-state.adoc

@@ -1,56 +1,84 @@
 [#0794-valid-tic-tac-toe-state]
-= 794. Valid Tic-Tac-Toe State
+= 794. 有效的井字游戏
 
-{leetcode}/problems/valid-tic-tac-toe-state/[LeetCode - Valid Tic-Tac-Toe State^]
+https://leetcode.cn/problems/valid-tic-tac-toe-state/[LeetCode - 794. 有效的井字游戏 ^]
 
-A Tic-Tac-Toe board is given 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.
+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` 。
 
-The `board` is a 3 x 3 array, and consists of characters `" "`, `"X"`, and `"O"`.  The " " character represents an empty square.
+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。
 
-Here are the rules of Tic-Tac-Toe:
+以下是井字游戏的规则：
 
+* 玩家轮流将字符放入空位（`' '`）中。
+* 玩家 1 总是放字符 `'X'` ，而玩家 2 总是放字符 `'O'` 。
+* `'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。
+* 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。
+* 当所有位置非空时，也算为游戏结束。
+* 如果游戏结束，玩家不允许再放置字符。
 
-* Players take turns placing characters into empty squares (" ").
-* The first player always places "X" characters, while the second player always places "O" characters.
-* "X" and "O" characters are always placed into empty squares, never 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.
+*示例 1：*
 
+image::images/0794-01.jpg[{image_attr}]
 
-[subs="verbatim,quotes,macros"]
-----
-*Example 1:*
-*Input:* board = ["O  ", "   ", "   "]
-*Output:* false
-*Explanation:* The first player always plays "X".
-
-*Example 2:*
-*Input:* board = ["XOX", " X ", "   "]
-*Output:* false
-*Explanation:* Players take turns making moves.
-
-*Example 3:*
-*Input:* board = ["XXX", "   ", "OOO"]
-*Output:* false
-
-*Example 4:*
-*Input:* board = ["XOX", "O O", "XOX"]
-*Output:* true
-----
+....
+输入：board = ["O  ","   ","   "]
+输出：false
+解释：玩家 1 总是放字符 "X" 。
+....
+
+*示例 2：*
+
+image::images/0794-02.jpg[{image_attr}]
+
+....
+输入：board = ["XOX"," X ","   "]
+输出：false
+解释：玩家应该轮流放字符。
+....
+
+*示例 3:*
 
-*Note:*
+image::images/0794-03.jpg[{image_attr}]
 
+....
+输入：board = ["XOX","O O","XOX"]
+输出：true
+....
 
-* `board` is a length-3 array of strings, where each string `board[i]` has length 3.
-* Each `board[i][j]` is a character in the set `{" ", "X", "O"}`.
+*提示：*
 
+* `board.length == 3`
+* `board[i].length == 3`
+* `board[i][j]` 为 `'X'`、`'O'` 或 `' '`
 
 
 
+== 思路分析
+
+
 [[src-0794]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0794_ValidTicTacToeState.java[tag=answer]
 ----
+--
+
+// 二刷::
+// +
+// --
+// [{java_src_attr}]
+// ----
+// include::{sourcedir}/_0794_ValidTicTacToeState_2.java[tag=answer]
+// ----
+// --
+====
+
+
+== 参考资料
+
 

docs/images/0794-01.jpg

@@ -1672,7 +1672,7 @@ include::0787-cheapest-flights-within-k-stops.adoc[leveloffset=+1]
 
 // include::0793-preimage-size-of-factorial-zeroes-function.adoc[leveloffset=+1]
 
-// include::0794-valid-tic-tac-toe-state.adoc[leveloffset=+1]
+include::0794-valid-tic-tac-toe-state.adoc[leveloffset=+1]
 
 // include::0795-number-of-subarrays-with-bounded-maximum.adoc[leveloffset=+1]
 

docs/images/0794-02.jpg

@@ -618,6 +618,11 @@ endif::[]
 |{doc_base_url}/0787-cheapest-flights-within-k-stops.adoc[题解]
 |⭕️ 深度优先搜索通过 48 / 56 个测试用例。使用带备忘录的深度优先搜索顺利通过。带备忘录的深度优先搜索，也可以优化成动态规划。
 
+|{counter:codes2503}
+|{leetcode_base_url}/valid-tic-tac-toe-state/[794. 有效的井字游戏^]
+|{doc_base_url}/0794-valid-tic-tac-toe-state.adoc[题解]
+|✅ 总结提炼判断规则。
+
 
 |===
 

docs/images/0794-03.jpg

@@ -0,0 +1,72 @@
+package com.diguage.algo.leetcode;
+
+public class _0794_ValidTicTacToeState {
+  // tag::answer[]
+
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-05-07 21:52:04
+   */
+  public boolean validTicTacToe(String[] board) {
+    int length = board.length;
+    char[][] grid = new char[length][length];
+    int xc = 0, oc = 0;
+    boolean xRowOver = false;
+    boolean oRowOver = false;
+    for (int row = 0; row < length; row++) {
+      char[] chars = board[row].toCharArray();
+      int ixc = 0;
+      int ioc = 0;
+      for (int column = 0; column < chars.length; column++) {
+        char c = chars[column];
+        if (c == 'X') {
+          xc++;
+          ixc++;
+        } else if (c == 'O') {
+          oc++;
+          ioc++;
+        }
+        grid[row][column] = c;
+      }
+      if (!xRowOver && ixc == 3) {
+        xRowOver = true;
+      }
+      if (!oRowOver && ioc == 3) {
+        oRowOver = true;
+      }
+    }
+    // X 必须比 O 多，但是最多能多一个
+    // O 不会比 X 多
+    if (xc - oc > 1 || oc - xc >= 1) {
+      return false;
+    }
+    boolean xColOver = false;
+    boolean oColOver = false;
+    for (int i = 0; i < length; i++) {
+      xColOver = grid[0][i] == 'X' && grid[1][i] == 'X' && grid[2][i] == 'X';
+      oColOver = grid[0][i] == 'O' && grid[1][i] == 'O' && grid[2][i] == 'O';
+      if (xColOver || oColOver) {
+        break;
+      }
+    }
+    boolean xDiagOver = (grid[0][0] == 'X' && grid[1][1] == 'X' && grid[2][2] == 'X')
+      || (grid[0][2] == 'X' && grid[1][1] == 'X' && grid[2][0] == 'X');
+    boolean oDiagOver = (grid[0][0] == 'O' && grid[1][1] == 'O' && grid[2][2] == 'O')
+      || (grid[0][2] == 'O' && grid[1][1] == 'O' && grid[2][0] == 'O');
+    // 如果 X 获胜，则 X 比 O 多一个
+    if ((xRowOver || xColOver || xDiagOver) && xc - oc != 1) {
+      return false;
+    }
+    // 如果 O 获胜，则 X 与 O 个数相等
+    if ((oRowOver || oColOver || oDiagOver) && xc != oc) {
+      return false;
+    }
+    return true;
+  }
+
+  // end::answer[]
+  public static void main(String[] args) {
+    new _0794_ValidTicTacToeState()
+      .validTicTacToe(new String[]{"OXX", "XOX", "OXO"});
+  }
+}