feat: add solutions to leetcode problem: No.1275. Find Winner on a Tic Tac Toe Game

Author: yanglbme

solution/1200-1299/1275.Find Winner on a Tic Tac Toe Game/README.md

@@ -79,27 +79,85 @@
 	<li><code>moves</code> 遵循井字棋的规则。</li>
 </ul>
 
-
 ## 解法
 
 <!-- 这里可写通用的实现逻辑 -->
 
+判断 A、B 谁能获胜，只需判断最后一个落棋的人能否获胜即可。我们用数组 `counter` 记录 `0~2` 行、`0~2` 列、`正对角线`、`副对角线`是否已满 3 个棋子。如果等于 3，此人获胜，游戏结束。
+
+若最后落棋者为未能获胜，棋盘被下满返回 `Draw`，未下满则返回 `Pending`。
+
+> 数组 `counter` 长度为 8，其中，`counter[0..2]` 对应 `0~2` 行，`counter[3..5]` 对应 `0~2` 列，`counter[6]` 对应正对角线，`counter[7]` 对应副对角线的落棋次数。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def tictactoe(self, moves: List[List[int]]) -> str:
+        n = len(moves)
+        counter = [0] * 8
+        for i in range(n - 1, -1, -2):
+            row, col = moves[i][0], moves[i][1]
+            counter[row] += 1
+            counter[col + 3] += 1
+            if row == col:
+                counter[6] += 1
+            if row + col == 2:
+                counter[7] += 1
+            if counter[row] == 3 or counter[col + 3] == 3 or counter[6] == 3 or counter[7] == 3:
+                return "A" if (i % 2) == 0 else "B"
+        return "Draw" if n == 9 else "Pending"
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
+class Solution {
+    public String tictactoe(int[][] moves) {
+        int n = moves.length;
+        int[] counter = new int[8];
+        for (int i = n - 1; i >= 0; i -= 2) {
+            int row = moves[i][0], col = moves[i][1];
+            ++counter[row];
+            ++counter[col + 3];
+            if (row == col) ++counter[6];
+            if (row + col == 2) ++counter[7];
+            if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
+                return (i % 2) == 0 ? "A" : "B";
+            }
+        }
+        return n == 9 ? "Draw" : "Pending";
+    }
+}
+```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string tictactoe(vector<vector<int>>& moves) {
+        int n = moves.size();
+        vector<int> counter(8, 0);
+        for (int i = n - 1; i >= 0; i -= 2) {
+            int row = moves[i][0], col = moves[i][1];
+            ++counter[row];
+            ++counter[col + 3];
+            if (row == col) ++counter[6];
+            if (row + col == 2) ++counter[7];
+            if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
+                return (i % 2 == 0) ? "A" : "B";
+            }
+        }
+        return n == 9 ? "Draw" : "Pending";
+    }
+};
 ```
 
 ### **...**

solution/1200-1299/1275.Find Winner on a Tic Tac Toe Game/README_EN.md

@@ -6,12 +6,8 @@
 
 <p>Tic-tac-toe is played&nbsp;by&nbsp;two players <em>A</em> and <em>B</em> on a&nbsp;<i>3</i>&nbsp;x&nbsp;<i>3</i>&nbsp;grid.</p>
 
-
-
 <p>Here are the rules of Tic-Tac-Toe:</p>
 
-
-
 <ul>
 	<li>Players take turns placing characters into empty squares (&quot; &quot;).</li>
 	<li>The first player <em>A</em> always places &quot;X&quot; characters, while the second player <em>B</em>&nbsp;always places &quot;O&quot; characters.</li>
@@ -21,26 +17,16 @@
 	<li>No more moves can be played if the game is over.</li>
 </ul>
 
-
-
 <p>Given an array <code>moves</code> where each element&nbsp;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 <em>A</em> and <em>B</em> play.</p>
 
-
-
 <p>Return the winner of the game if it exists (<em>A</em> or <em>B</em>), in case the game ends in a draw return &quot;Draw&quot;, if there are still movements to play return &quot;Pending&quot;.</p>
 
-
-
 <p>You can assume that&nbsp;<code>moves</code> is&nbsp;<strong>valid</strong> (It follows the rules of Tic-Tac-Toe),&nbsp;the grid is initially empty and <em>A</em> will play <strong>first</strong>.</p>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Example 1:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> moves = [[0,0],[2,0],[1,1],[2,1],[2,2]]
@@ -57,12 +43,8 @@
 
 </pre>
 
-
-
 <p><strong>Example 2:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> moves = [[0,0],[1,1],[0,1],[0,2],[1,0],[2,0]]
@@ -79,12 +61,8 @@
 
 </pre>
 
-
-
 <p><strong>Example 3:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> moves = [[0,0],[1,1],[2,0],[1,0],[1,2],[2,1],[0,1],[0,2],[2,2]]
@@ -101,12 +79,8 @@
 
 </pre>
 
-
-
 <p><strong>Example 4:</strong></p>
 
-
-
 <pre>
 
 <strong>Input:</strong> moves = [[0,0],[1,1]]
@@ -123,14 +97,10 @@
 
 </pre>
 
-
-
 <p>&nbsp;</p>
 
 <p><strong>Constraints:</strong></p>
 
-
-
 <ul>
 	<li><code>1 &lt;= moves.length &lt;= 9</code></li>
 	<li><code>moves[i].length == 2</code></li>
@@ -146,13 +116,66 @@
 ### **Python3**
 
 ```python
-
+class Solution:
+    def tictactoe(self, moves: List[List[int]]) -> str:
+        n = len(moves)
+        counter = [0] * 8
+        for i in range(n - 1, -1, -2):
+            row, col = moves[i][0], moves[i][1]
+            counter[row] += 1
+            counter[col + 3] += 1
+            if row == col:
+                counter[6] += 1
+            if row + col == 2:
+                counter[7] += 1
+            if counter[row] == 3 or counter[col + 3] == 3 or counter[6] == 3 or counter[7] == 3:
+                return "A" if (i % 2) == 0 else "B"
+        return "Draw" if n == 9 else "Pending"
 ```
 
 ### **Java**
 
 ```java
+class Solution {
+    public String tictactoe(int[][] moves) {
+        int n = moves.length;
+        int[] counter = new int[8];
+        for (int i = n - 1; i >= 0; i -= 2) {
+            int row = moves[i][0], col = moves[i][1];
+            ++counter[row];
+            ++counter[col + 3];
+            if (row == col) ++counter[6];
+            if (row + col == 2) ++counter[7];
+            if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
+                return (i % 2) == 0 ? "A" : "B";
+            }
+        }
+        return n == 9 ? "Draw" : "Pending";
+    }
+}
+```
 
+### **C++**
+
+```cpp
+class Solution {
+public:
+    string tictactoe(vector<vector<int>>& moves) {
+        int n = moves.size();
+        vector<int> counter(8, 0);
+        for (int i = n - 1; i >= 0; i -= 2) {
+            int row = moves[i][0], col = moves[i][1];
+            ++counter[row];
+            ++counter[col + 3];
+            if (row == col) ++counter[6];
+            if (row + col == 2) ++counter[7];
+            if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
+                return (i % 2 == 0) ? "A" : "B";
+            }
+        }
+        return n == 9 ? "Draw" : "Pending";
+    }
+};
 ```
 
 ### **...**

solution/1200-1299/1275.Find Winner on a Tic Tac Toe Game/Solution.cpp

@@ -1,30 +1,18 @@
 class Solution {
-    bool checkWin(vector<vector<char>>& board, bool moveA) {
-        char symbol = (moveA ? 'X' : 'O');
-        for (int i = 0; i < 3; i++)
-            if (board[i][0] == symbol && board[i][0] == board[i][1] && board[i][0] == board[i][2]) 
-                return true;
-        for (int i = 0; i < 3; i++)
-            if (board[0][i] == symbol && board[0][i] == board[1][i] && board[0][i] == board[2][i])
-                return true;
-        if (board[0][0] == symbol && board[1][1] == symbol && board[2][2] == symbol)
-            return true;
-        if (board[0][2] == symbol && board[1][1] == symbol && board[2][0] == symbol)
-            return true;
-        return false;
-    }
 public:
     string tictactoe(vector<vector<int>>& moves) {
-        vector<vector<char>> board(3, vector<char>(3, ' '));
-        bool moveA = true;
-        for (auto &v : moves) {
-            board[v[0]][v[1]] = (moveA ? 'X' : 'O');
-            if (checkWin(board, moveA))
-                return (moveA ? "A" : "B");
-            moveA = !moveA;
+        int n = moves.size();
+        vector<int> counter(8, 0);
+        for (int i = n - 1; i >= 0; i -= 2) {
+            int row = moves[i][0], col = moves[i][1];
+            ++counter[row];
+            ++counter[col + 3];
+            if (row == col) ++counter[6];
+            if (row + col == 2) ++counter[7];
+            if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
+                return (i % 2 == 0) ? "A" : "B";
+            }
         }
-        if (moves.size() == 9)
-            return "Draw";
-        return "Pending";
+        return n == 9 ? "Draw" : "Pending";
     }
 };

solution/1200-1299/1275.Find Winner on a Tic Tac Toe Game/Solution.java

@@ -0,0 +1,17 @@
+class Solution {
+    public String tictactoe(int[][] moves) {
+        int n = moves.length;
+        int[] counter = new int[8];
+        for (int i = n - 1; i >= 0; i -= 2) {
+            int row = moves[i][0], col = moves[i][1];
+            ++counter[row];
+            ++counter[col + 3];
+            if (row == col) ++counter[6];
+            if (row + col == 2) ++counter[7];
+            if (counter[row] == 3 || counter[col + 3] == 3 || counter[6] == 3 || counter[7] == 3) {
+                return (i % 2) == 0 ? "A" : "B";
+            }
+        }
+        return n == 9 ? "Draw" : "Pending";
+    }
+}

solution/1200-1299/1275.Find Winner on a Tic Tac Toe Game/Solution.py

@@ -0,0 +1,15 @@
+class Solution:
+    def tictactoe(self, moves: List[List[int]]) -> str:
+        n = len(moves)
+        counter = [0] * 8
+        for i in range(n - 1, -1, -2):
+            row, col = moves[i][0], moves[i][1]
+            counter[row] += 1
+            counter[col + 3] += 1
+            if row == col:
+                counter[6] += 1
+            if row + col == 2:
+                counter[7] += 1
+            if counter[row] == 3 or counter[col + 3] == 3 or counter[6] == 3 or counter[7] == 3:
+                return "A" if (i % 2) == 0 else "B"
+        return "Draw" if n == 9 else "Pending"