feat: update lc problems

solution/0500-0599/0552.Student Attendance Record II/README.md

@@ -152,36 +152,30 @@ class Solution {
 #### C++
 
 ```cpp
-int f[100010][2][3];
-const int mod = 1e9 + 7;
-
 class Solution {
 public:
     int checkRecord(int n) {
-        this->n = n;
+        int f[n][2][3];
         memset(f, -1, sizeof(f));
-        return dfs(0, 0, 0);
-    }
-
-    int dfs(int i, int j, int k) {
-        if (i >= n) {
-            return 1;
-        }
-        if (f[i][j][k] != -1) {
-            return f[i][j][k];
-        }
-        int ans = dfs(i + 1, j, 0);
-        if (j == 0) {
-            ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
-        }
-        if (k < 2) {
-            ans = (ans + dfs(i + 1, j, k + 1)) % mod;
-        }
-        return f[i][j][k] = ans;
+        const int mod = 1e9 + 7;
+        auto dfs = [&](auto&& dfs, int i, int j, int k) -> int {
+            if (i >= n) {
+                return 1;
+            }
+            if (f[i][j][k] != -1) {
+                return f[i][j][k];
+            }
+            int ans = dfs(dfs, i + 1, j, 0);
+            if (j == 0) {
+                ans = (ans + dfs(dfs, i + 1, j + 1, 0)) % mod;
+            }
+            if (k < 2) {
+                ans = (ans + dfs(dfs, i + 1, j, k + 1)) % mod;
+            }
+            return f[i][j][k] = ans;
+        };
+        return dfs(dfs, 0, 0, 0);
     }
-
-private:
-    int n;
 };
 ```
 

solution/0500-0599/0552.Student Attendance Record II/README_EN.md

@@ -134,36 +134,30 @@ class Solution {
 #### C++
 
 ```cpp
-int f[100010][2][3];
-const int mod = 1e9 + 7;
-
 class Solution {
 public:
     int checkRecord(int n) {
-        this->n = n;
+        int f[n][2][3];
         memset(f, -1, sizeof(f));
-        return dfs(0, 0, 0);
-    }
-
-    int dfs(int i, int j, int k) {
-        if (i >= n) {
-            return 1;
-        }
-        if (f[i][j][k] != -1) {
-            return f[i][j][k];
-        }
-        int ans = dfs(i + 1, j, 0);
-        if (j == 0) {
-            ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
-        }
-        if (k < 2) {
-            ans = (ans + dfs(i + 1, j, k + 1)) % mod;
-        }
-        return f[i][j][k] = ans;
+        const int mod = 1e9 + 7;
+        auto dfs = [&](auto&& dfs, int i, int j, int k) -> int {
+            if (i >= n) {
+                return 1;
+            }
+            if (f[i][j][k] != -1) {
+                return f[i][j][k];
+            }
+            int ans = dfs(dfs, i + 1, j, 0);
+            if (j == 0) {
+                ans = (ans + dfs(dfs, i + 1, j + 1, 0)) % mod;
+            }
+            if (k < 2) {
+                ans = (ans + dfs(dfs, i + 1, j, k + 1)) % mod;
+            }
+            return f[i][j][k] = ans;
+        };
+        return dfs(dfs, 0, 0, 0);
     }
-
-private:
-    int n;
 };
 ```
 

solution/0500-0599/0552.Student Attendance Record II/Solution.cpp

@@ -1,31 +1,25 @@
-int f[100010][2][3];
-const int mod = 1e9 + 7;
-
 class Solution {
 public:
     int checkRecord(int n) {
-        this->n = n;
+        int f[n][2][3];
         memset(f, -1, sizeof(f));
-        return dfs(0, 0, 0);
+        const int mod = 1e9 + 7;
+        auto dfs = [&](auto&& dfs, int i, int j, int k) -> int {
+            if (i >= n) {
+                return 1;
+            }
+            if (f[i][j][k] != -1) {
+                return f[i][j][k];
+            }
+            int ans = dfs(dfs, i + 1, j, 0);
+            if (j == 0) {
+                ans = (ans + dfs(dfs, i + 1, j + 1, 0)) % mod;
+            }
+            if (k < 2) {
+                ans = (ans + dfs(dfs, i + 1, j, k + 1)) % mod;
+            }
+            return f[i][j][k] = ans;
+        };
+        return dfs(dfs, 0, 0, 0);
     }
-
-    int dfs(int i, int j, int k) {
-        if (i >= n) {
-            return 1;
-        }
-        if (f[i][j][k] != -1) {
-            return f[i][j][k];
-        }
-        int ans = dfs(i + 1, j, 0);
-        if (j == 0) {
-            ans = (ans + dfs(i + 1, j + 1, 0)) % mod;
-        }
-        if (k < 2) {
-            ans = (ans + dfs(i + 1, j, k + 1)) % mod;
-        }
-        return f[i][j][k] = ans;
-    }
-
-private:
-    int n;
-};
+};

solution/0700-0799/0756.Pyramid Transition Matrix/README.md

@@ -40,8 +40,8 @@ tags:
 <strong>输入：</strong>bottom = "BCD", allowed = ["BCC","CDE","CEA","FFF"]
 <strong>输出：</strong>true
 <strong>解释：</strong>允许的三角形模式显示在右边。
-从最底层(第3层)开始，我们可以在第2层构建“CE”，然后在第1层构建“E”。
-金字塔中有三种三角形图案，分别是“BCC”、“CDE”和“CEA”。都是允许的。
+从最底层(第 3 层)开始，我们可以在第 2 层构建“CE”，然后在第 1 层构建“E”。
+金字塔中有三种三角形图案，分别是 “BCC”、“CDE” 和 “CEA”。都是允许的。
 </pre>
 
 <p><strong>示例 2：</strong></p>
@@ -52,7 +52,7 @@ tags:
 <strong>输入：</strong>bottom = "AAAA", allowed = ["AAB","AAC","BCD","BBE","DEF"]
 <strong>输出：</strong>false
 <strong>解释：</strong>允许的三角形模式显示在右边。
-从最底层(游戏邦注:即第4个关卡)开始，创造第3个关卡有多种方法，但如果尝试所有可能性，你便会在创造第1个关卡前陷入困境。
+从最底层(游戏邦注:即第 4 个关卡)开始，创造第 3 个关卡有多种方法，但如果尝试所有可能性，你便会在创造第 1 个关卡前陷入困境。
 </pre>
 
 <p>&nbsp;</p>
@@ -63,7 +63,7 @@ tags:
 	<li><code>2 &lt;= bottom.length &lt;= 6</code></li>
 	<li><code>0 &lt;= allowed.length &lt;= 216</code></li>
 	<li><code>allowed[i].length == 3</code></li>
-	<li>所有输入字符串中的字母来自集合&nbsp;<code>{'A', 'B', 'C', 'D', 'E', 'F', 'G'}</code>。</li>
+	<li>所有输入字符串中的字母来自集合&nbsp;<code>{'A', 'B', 'C', 'D', 'E', 'F'}</code>。</li>
 	<li>&nbsp;<code>allowed</code>&nbsp;中所有值都是 <strong>唯一的</strong></li>
 </ul>
 

solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/README.md

@@ -314,7 +314,7 @@ function getDirections(root: TreeNode | null, startValue: number, destValue: num
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {
@@ -370,7 +370,7 @@ var getDirections = function (root, startValue, destValue) {
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node, x, path) => {

solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/README_EN.md

@@ -310,7 +310,7 @@ function getDirections(root: TreeNode | null, startValue: number, destValue: num
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {
@@ -366,7 +366,7 @@ var getDirections = function (root, startValue, destValue) {
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node, x, path) => {

solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/Solution.js

@@ -20,7 +20,7 @@ var getDirections = function (root, startValue, destValue) {
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node, x, path) => {

solution/2000-2099/2096.Step-By-Step Directions From a Binary Tree Node to Another/Solution.ts

@@ -20,7 +20,7 @@ function getDirections(root: TreeNode | null, startValue: number, destValue: num
         const left = lca(node.left, p, q);
         const right = lca(node.right, p, q);
 
-        return left && right ? node : left ?? right;
+        return left && right ? node : (left ?? right);
     };
 
     const dfs = (node: TreeNode | null, x: number, path: string[]): boolean => {

solution/3100-3199/3154.Find Number of Ways to Reach the K-th Stair/README.md

@@ -187,29 +187,24 @@ class Solution {
 class Solution {
 public:
     int waysToReachStair(int k) {
-        this->k = k;
-        return dfs(1, 0, 0);
-    }
-
-private:
-    unordered_map<long long, int> f;
-    int k;
-
-    int dfs(int i, int j, int jump) {
-        if (i > k + 1) {
-            return 0;
-        }
-        long long key = ((long long) i << 32) | jump << 1 | j;
-        if (f.contains(key)) {
-            return f[key];
-        }
-        int ans = i == k ? 1 : 0;
-        if (i > 0 && j == 0) {
-            ans += dfs(i - 1, 1, jump);
-        }
-        ans += dfs(i + (1 << jump), 0, jump + 1);
-        f[key] = ans;
-        return ans;
+        unordered_map<long long, int> f;
+        auto dfs = [&](auto&& dfs, int i, int j, int jump) -> int {
+            if (i > k + 1) {
+                return 0;
+            }
+            long long key = ((long long) i << 32) | jump << 1 | j;
+            if (f.contains(key)) {
+                return f[key];
+            }
+            int ans = i == k ? 1 : 0;
+            if (i > 0 && j == 0) {
+                ans += dfs(dfs, i - 1, 1, jump);
+            }
+            ans += dfs(dfs, i + (1 << jump), 0, jump + 1);
+            f[key] = ans;
+            return ans;
+        };
+        return dfs(dfs, 1, 0, 0);
     }
 };
 ```

solution/3100-3199/3154.Find Number of Ways to Reach the K-th Stair/README_EN.md

@@ -185,29 +185,24 @@ class Solution {
 class Solution {
 public:
     int waysToReachStair(int k) {
-        this->k = k;
-        return dfs(1, 0, 0);
-    }
-
-private:
-    unordered_map<long long, int> f;
-    int k;
-
-    int dfs(int i, int j, int jump) {
-        if (i > k + 1) {
-            return 0;
-        }
-        long long key = ((long long) i << 32) | jump << 1 | j;
-        if (f.contains(key)) {
-            return f[key];
-        }
-        int ans = i == k ? 1 : 0;
-        if (i > 0 && j == 0) {
-            ans += dfs(i - 1, 1, jump);
-        }
-        ans += dfs(i + (1 << jump), 0, jump + 1);
-        f[key] = ans;
-        return ans;
+        unordered_map<long long, int> f;
+        auto dfs = [&](auto&& dfs, int i, int j, int jump) -> int {
+            if (i > k + 1) {
+                return 0;
+            }
+            long long key = ((long long) i << 32) | jump << 1 | j;
+            if (f.contains(key)) {
+                return f[key];
+            }
+            int ans = i == k ? 1 : 0;
+            if (i > 0 && j == 0) {
+                ans += dfs(dfs, i - 1, 1, jump);
+            }
+            ans += dfs(dfs, i + (1 << jump), 0, jump + 1);
+            f[key] = ans;
+            return ans;
+        };
+        return dfs(dfs, 1, 0, 0);
     }
 };
 ```

solution/3100-3199/3154.Find Number of Ways to Reach the K-th Stair/Solution.cpp

@@ -1,28 +1,23 @@
 class Solution {
 public:
     int waysToReachStair(int k) {
-        this->k = k;
-        return dfs(1, 0, 0);
+        unordered_map<long long, int> f;
+        auto dfs = [&](auto&& dfs, int i, int j, int jump) -> int {
+            if (i > k + 1) {
+                return 0;
+            }
+            long long key = ((long long) i << 32) | jump << 1 | j;
+            if (f.contains(key)) {
+                return f[key];
+            }
+            int ans = i == k ? 1 : 0;
+            if (i > 0 && j == 0) {
+                ans += dfs(dfs, i - 1, 1, jump);
+            }
+            ans += dfs(dfs, i + (1 << jump), 0, jump + 1);
+            f[key] = ans;
+            return ans;
+        };
+        return dfs(dfs, 1, 0, 0);
     }
-
-private:
-    unordered_map<long long, int> f;
-    int k;
-
-    int dfs(int i, int j, int jump) {
-        if (i > k + 1) {
-            return 0;
-        }
-        long long key = ((long long) i << 32) | jump << 1 | j;
-        if (f.contains(key)) {
-            return f[key];
-        }
-        int ans = i == k ? 1 : 0;
-        if (i > 0 && j == 0) {
-            ans += dfs(i - 1, 1, jump);
-        }
-        ans += dfs(i + (1 << jump), 0, jump + 1);
-        f[key] = ans;
-        return ans;
-    }
-};
+};