update: 添加问题“840.矩阵中的幻方”的代码和题解 (#1293)

LetMeFly666 · web-flow · commit b24bdcabce0a · 2025-12-30T18:59:05.000+08:00
* word: en (#1292) * 849: WA.cpp (#1292) * 840: AC.cpp+WA.py (#1292) cpp - AC,100.00%,69.12% * 840: AC.py (#1292) - AC,23.81%,100.00% * 840: half.题解 (#1292) - 憩 * update: 添加问题“840.矩阵中的幻方”的代码和题解 (#1293) Signed-off-by: LetMeFly666 <814114971@qq.com> --------- Signed-off-by: LetMeFly666 <814114971@qq.com>
diff --git a/Codes/0840-magic-squares-in-grid.cpp b/Codes/0840-magic-squares-in-grid.cpp
@@ -0,0 +1,80 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2025-12-30 13:14:11
+ * @LastEditors: LetMeFly.xyz
+ * @LastEditTime: 2025-12-30 13:25:00
+ */
+#if defined(_WIN32) || defined(__APPLE__)
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+private:
+    inline bool is(vector<vector<int>>& grid, int i, int j) {
+        int mask = 0;
+        int rowCnt[3] = {0}, colCnt[3] = {0};
+        for (int di = 0; di < 3; di++) {
+            for (int dj = 0; dj < 3; dj++) {
+                int v = grid[i - di][j - dj];
+                mask |= 1 << v;
+                rowCnt[di] += v;
+                colCnt[dj] += v;
+            }
+        }
+        if (mask != (1 << 10) - 2) {  // (1<<10)-1：1111111111（10个1）而mask没有或上1<<0所以再-1
+            return false;
+        }
+
+        int cnt = rowCnt[0];
+        for (int d = 0; d < 3; d++) {
+            if (rowCnt[d] != cnt || colCnt[d] != cnt) {
+                return false;
+            }
+        }
+        if (grid[i][j] + grid[i - 1][j - 1] + grid[i - 2][j - 2] != cnt) {
+            return false;
+        }
+        if (grid[i - 2][j] + grid[i - 1][j - 1] + grid[i][j - 2] != cnt) {
+            return false;
+        }
+        return true;
+    }
+public:
+    int numMagicSquaresInside(vector<vector<int>>& grid) {
+        int ans = 0;
+        int n = grid.size(), m = grid[0].size();
+        for (int i = 2; i < n; i++) {
+            for (int j = 2; j < m; j++) {
+                ans += is(grid, i, j);
+            }
+        }
+        return ans;
+    }
+};
+
+#if defined(_WIN32) || defined(__APPLE__)
+/*
+[[4,3,8,4],[9,5,1,9],[2,7,6,2]]
+
+1
+*/
+void dbg1to9() {
+    int mask = 0;
+    for (int i = 1; i < 10; i++) {
+        mask |= 1 << i;
+    }
+    cout << mask << endl;  // 1022不是1023，因为mask没有或上1<<0
+}
+
+int main() {
+    dbg1to9();
+
+    string s;
+    while (cin >> s) {
+        Solution sol;
+        vector<vector<int>> v = stringToVectorVector(s);
+        cout << sol.numMagicSquaresInside(v) << endl;
+    }
+    return 0;
+}
+#endif
diff --git a/Codes/0840-magic-squares-in-grid.py b/Codes/0840-magic-squares-in-grid.py
@@ -0,0 +1,32 @@
+'''
+Author: LetMeFly
+Date: 2025-12-30 13:14:11
+LastEditors: LetMeFly.xyz
+LastEditTime: 2025-12-30 13:38:58
+'''
+from typing import List
+
+class Solution:
+    def ok(self, grid: List[List[int]], i: int, j: int) -> bool:
+        mask = 0
+        colCnt = [0] * 3
+        rowCnt = [0] * 3
+        for di in range(3):
+            for dj in range(3):
+                v = grid[i - di][j - dj]
+                mask |= 1 << v
+                rowCnt[di] += v
+                colCnt[dj] += v
+        if mask != (1 << 10) - 2:
+            return False
+        cnt = grid[i][j] + grid[i - 1][j - 1] + grid[i - 2][j - 2]
+        if grid[i - 2][j] + grid[i - 1][j - 1] + grid[i][j - 2] != cnt:
+            return False
+        if any(c != cnt for c in colCnt):
+            return False
+        if any(r != cnt for r in rowCnt):
+            return False
+        return True
+    
+    def numMagicSquaresInside(self, grid: List[List[int]]) -> int:
+        return sum(self.ok(grid, i, j) for j in range(2, len(grid[0])) for i in range(2, len(grid)))
diff --git a/README.md b/README.md
@@ -432,6 +432,7 @@
 |0831.隐藏个人信息|中等|<a href="https://leetcode.cn/problems/masking-personal-information/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/04/01/LeetCode%200831.%E9%9A%90%E8%97%8F%E4%B8%AA%E4%BA%BA%E4%BF%A1%E6%81%AF/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/129893329" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/masking-personal-information/solutions/2202064/letmefly-831yin-cang-ge-ren-xin-xi-by-ti-xkxe/" target="_blank">LeetCode题解</a>|
 |0833.字符串中的查找与替换|中等|<a href="https://leetcode.cn/problems/find-and-replace-in-string/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/08/15/LeetCode%200833.%E5%AD%97%E7%AC%A6%E4%B8%B2%E4%B8%AD%E7%9A%84%E6%9F%A5%E6%89%BE%E4%B8%8E%E6%9B%BF%E6%8D%A2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132289306" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/find-and-replace-in-string/solutions/2388863/letmefly-833zi-fu-chuan-zhong-de-cha-zha-f9vs/" target="_blank">LeetCode题解</a>|
 |0837.新21点|中等|<a href="https://leetcode.cn/problems/new-21-game/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/08/17/LeetCode%200837.%E6%96%B021%E7%82%B9/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/150474266" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/new-21-game/solutions/3755850/letmefly-837xin-21-dian-dong-tai-gui-hua-8o4t/" target="_blank">LeetCode题解</a>|
+|0840.矩阵中的幻方|中等|<a href="https://leetcode.cn/problems/magic-squares-in-grid/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2025/12/30/LeetCode%200840.%E7%9F%A9%E9%98%B5%E4%B8%AD%E7%9A%84%E5%B9%BB%E6%96%B9/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/156429151" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/magic-squares-in-grid/solutions/3869464/letmefly-840ju-zhen-zhong-de-huan-fang-m-m5ev/" target="_blank">LeetCode题解</a>|
 |0849.到最近的人的最大距离|中等|<a href="https://leetcode.cn/problems/maximize-distance-to-closest-person/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/08/22/LeetCode%200849.%E5%88%B0%E6%9C%80%E8%BF%91%E7%9A%84%E4%BA%BA%E7%9A%84%E6%9C%80%E5%A4%A7%E8%B7%9D%E7%A6%BB/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/132420545" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximize-distance-to-closest-person/solutions/2399151/letmefly-849dao-zui-jin-de-ren-de-zui-da-dnw3/" target="_blank">LeetCode题解</a>|
 |0856.括号的分数|中等|<a href="https://leetcode.cn/problems/score-of-parentheses/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/09/LeetCode%200856.%E6%8B%AC%E5%8F%B7%E7%9A%84%E5%88%86%E6%95%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127221656" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/score-of-parentheses/solution/letmefly-856gua-hao-de-fen-shu-by-tisfy-v01a/" target="_blank">LeetCode题解</a>|
 |0860.柠檬水找零|简单|<a href="https://leetcode.cn/problems/lemonade-change/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/07/22/LeetCode%200860.%E6%9F%A0%E6%AA%AC%E6%B0%B4%E6%89%BE%E9%9B%B6/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131864033" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/lemonade-change/solutions/2353759/letmefly-860ning-meng-shui-zhao-ling-fu-5v6m1/" target="_blank">LeetCode题解</a>|
diff --git a/Solutions/LeetCode 0840.矩阵中的幻方.md b/Solutions/LeetCode 0840.矩阵中的幻方.md
@@ -0,0 +1,157 @@
+---
+title: 840.矩阵中的幻方：模拟（+小小位运算）
+date: 2025-12-30 13:40:02
+tags: [题解, LeetCode, 中等, 数组, 哈希表, 数学, 矩阵, 模拟, 位运算]
+categories: [题解, LeetCode]
+---
+
+# 【LetMeFly】840.矩阵中的幻方：模拟（+小小位运算）
+
+力扣题目链接：[https://leetcode.cn/problems/magic-squares-in-grid/](https://leetcode.cn/problems/magic-squares-in-grid/)
+
+<p><code>3 x 3</code> 的幻方是一个填充有&nbsp;<strong>从 <code>1</code> 到 <code>9</code>&nbsp;</strong> 的不同数字的 <code>3 x 3</code> 矩阵，其中每行，每列以及两条对角线上的各数之和都相等。</p>
+
+<p>给定一个由整数组成的<code>row x col</code>&nbsp;的 <code>grid</code>，其中有多少个&nbsp;<code>3 × 3</code> 的 “幻方” 子矩阵？</p>
+
+<p>注意：虽然幻方只能包含 1 到 9 的数字，但&nbsp;<code>grid</code> 可以包含最多15的数字。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+
+<p><img src="https://assets.leetcode.com/uploads/2020/09/11/magic_main.jpg" /></p>
+
+<pre>
+<strong>输入: </strong>grid = [[4,3,8,4],[9,5,1,9],[2,7,6,2]
+<strong>输出: </strong>1
+<strong>解释: </strong>
+下面的子矩阵是一个 3 x 3 的幻方：
+<img src="https://assets.leetcode.com/uploads/2020/09/11/magic_valid.jpg" />
+而这一个不是：
+<img src="https://assets.leetcode.com/uploads/2020/09/11/magic_invalid.jpg" />
+总的来说，在本示例所给定的矩阵中只有一个 3 x 3 的幻方子矩阵。
+</pre>
+
+<p><strong>示例 2:</strong></p>
+
+<pre>
+<strong>输入:</strong> grid = [[8]]
+<strong>输出:</strong> 0
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示:</strong></p>
+
+<ul>
+	<li><code>row == grid.length</code></li>
+	<li><code>col == grid[i].length</code></li>
+	<li><code>1 &lt;= row, col &lt;= 10</code></li>
+	<li><code>0 &lt;= grid[i][j] &lt;= 15</code></li>
+</ul>
+
+
+    
+## 解题方法：模拟
+
+主函数中枚举每个`3x3`矩阵的右下角下表，写一个辅助函数计算右下角坐标为`(i, j)`的矩阵是否是`幻方`。
+
+这个函数怎么写呢？
+
+对于是否由1-9组成，可以使用位运算，将所有数与初始值为0的mask按位或，如出现3则$mask |= 1 << 3$，最终看$mask$是否为$1<<10-2$。
+
+我们还可以使用两个大小为3的数组分别记录每一行和每一列的和，看他们是否相等、以及是否和主对角线和副对角线的和相等。
+
++ 时间复杂度$O(col\times row)$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+/*
+ * @LastEditTime: 2025-12-30 13:25:00
+ */
+class Solution {
+private:
+    inline bool is(vector<vector<int>>& grid, int i, int j) {
+        int mask = 0;
+        int rowCnt[3] = {0}, colCnt[3] = {0};
+        for (int di = 0; di < 3; di++) {
+            for (int dj = 0; dj < 3; dj++) {
+                int v = grid[i - di][j - dj];
+                mask |= 1 << v;
+                rowCnt[di] += v;
+                colCnt[dj] += v;
+            }
+        }
+        if (mask != (1 << 10) - 2) {  // (1<<10)-1：1111111111（10个1）而mask没有或上1<<0所以再-1
+            return false;
+        }
+
+        int cnt = rowCnt[0];
+        for (int d = 0; d < 3; d++) {
+            if (rowCnt[d] != cnt || colCnt[d] != cnt) {
+                return false;
+            }
+        }
+        if (grid[i][j] + grid[i - 1][j - 1] + grid[i - 2][j - 2] != cnt) {
+            return false;
+        }
+        if (grid[i - 2][j] + grid[i - 1][j - 1] + grid[i][j - 2] != cnt) {
+            return false;
+        }
+        return true;
+    }
+public:
+    int numMagicSquaresInside(vector<vector<int>>& grid) {
+        int ans = 0;
+        int n = grid.size(), m = grid[0].size();
+        for (int i = 2; i < n; i++) {
+            for (int j = 2; j < m; j++) {
+                ans += is(grid, i, j);
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Python
+
+```python
+'''
+LastEditTime: 2025-12-30 13:38:58
+'''
+from typing import List
+
+class Solution:
+    def ok(self, grid: List[List[int]], i: int, j: int) -> bool:
+        mask = 0
+        colCnt = [0] * 3
+        rowCnt = [0] * 3
+        for di in range(3):
+            for dj in range(3):
+                v = grid[i - di][j - dj]
+                mask |= 1 << v
+                rowCnt[di] += v
+                colCnt[dj] += v
+        if mask != (1 << 10) - 2:
+            return False
+        cnt = grid[i][j] + grid[i - 1][j - 1] + grid[i - 2][j - 2]
+        if grid[i - 2][j] + grid[i - 1][j - 1] + grid[i][j - 2] != cnt:
+            return False
+        if any(c != cnt for c in colCnt):
+            return False
+        if any(r != cnt for r in rowCnt):
+            return False
+        return True
+    
+    def numMagicSquaresInside(self, grid: List[List[int]]) -> int:
+        return sum(self.ok(grid, i, j) for j in range(2, len(grid[0])) for i in range(2, len(grid)))
+```
+
+> 同步发文于[CSDN](https://letmefly.blog.csdn.net/article/details/156429151)和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2025/12/30/LeetCode%200840.%E7%9F%A9%E9%98%B5%E4%B8%AD%E7%9A%84%E5%B9%BB%E6%96%B9/)哦~
+>
+> 千篇源码题解[已开源](https://github.com/LetMeFly666/LeetCode)
diff --git a/Solutions/Other-English-LearningNotes-SomeWords.md b/Solutions/Other-English-LearningNotes-SomeWords.md
@@ -1706,6 +1706,8 @@ categories: [自用]
 |||
 |menace|n. 威胁，危险的人/物，令人恐怖的氛围<br/>v. 对...构成危险，危及|
 |engraving|n. 版画，雕版印刷品，雕刻(术)|
+|||
+|ecstasy|n. 狂喜，陶醉，入迷；迷幻药|
 
 + 这个web要是能设计得可以闭眼(完全不睁眼)键盘控制背单词就好了。
 + 也许可以加个AI用最近词编故事功能(返回接口中支持标注所使用单词高亮?)
