feat: add solutions to lc problems: No.1215,1216

solution/1200-1299/1215.Stepping Numbers/README.md

@@ -34,6 +34,12 @@
 
 **方法一：BFS**
 
+首先，如果 $low$ 为 $0$，那么我们需要将 $0$ 加入答案中。
+
+接下来，我们创建一个队列 $q$，并将 $1 \sim 9$ 加入队列中。然后我们不断从队列中取出元素，记当前元素为 $v$，如果 $v$ 大于 $high$，那么我们就停止搜索；如果 $v$ 在 $[low, high]$ 的范围内，那么我们将 $v$ 加入答案中。然后，我们需要将 $v$ 的最后一位数字记为 $x$，如果 $x \gt 0$，那么我们将 $v \times 10 + x - 1$ 加入队列中；如果 $x \lt 9$，那么我们将 $v \times 10 + x + 1$ 加入队列中。重复上述操作，直到队列为空。
+
+时间复杂度 $O(10 \times 2^{\log M})$，空间复杂度 $O(2^{\log M})$，其中 $M$ 为 $high$ 的位数。
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -104,17 +110,29 @@ class Solution {
 public:
     vector<int> countSteppingNumbers(int low, int high) {
         vector<int> ans;
-        if (low == 0) ans.push_back(0);
+        if (low == 0) {
+            ans.push_back(0);
+        }
         queue<long long> q;
-        for (int i = 1; i < 10; ++i) q.push(i);
+        for (int i = 1; i < 10; ++i) {
+            q.push(i);
+        }
         while (!q.empty()) {
-            int v = q.front();
+            long long v = q.front();
             q.pop();
-            if (v > high) break;
-            if (v >= low) ans.push_back(v);
+            if (v > high) {
+                break;
+            }
+            if (v >= low) {
+                ans.push_back(v);
+            }
             int x = v % 10;
-            if (x) q.push(1ll * v * 10 + x - 1);
-            if (x < 9) q.push(1ll * v * 10 + x + 1);
+            if (x > 0) {
+                q.push(v * 10 + x - 1);
+            }
+            if (x < 9) {
+                q.push(v * 10 + x + 1);
+            }
         }
         return ans;
     }
@@ -151,6 +169,38 @@ func countSteppingNumbers(low int, high int) []int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function countSteppingNumbers(low: number, high: number): number[] {
+    const ans: number[] = [];
+    if (low === 0) {
+        ans.push(0);
+    }
+    const q: number[] = [];
+    for (let i = 1; i < 10; ++i) {
+        q.push(i);
+    }
+    while (q.length) {
+        const v = q.shift()!;
+        if (v > high) {
+            break;
+        }
+        if (v >= low) {
+            ans.push(v);
+        }
+        const x = v % 10;
+        if (x > 0) {
+            q.push(v * 10 + x - 1);
+        }
+        if (x < 9) {
+            q.push(v * 10 + x + 1);
+        }
+    }
+    return ans;
+}
+```
+
 ### **...**
 
 ```

solution/1200-1299/1215.Stepping Numbers/README_EN.md

@@ -36,6 +36,14 @@
 
 ## Solutions
 
+**Solution 1: BFS**
+
+First, if $low$ is $0$, we need to add $0$ to the answer.
+
+Next, we create a queue $q$ and add $1 \sim 9$ to the queue. Then, we repeatedly take out elements from the queue. Let the current element be $v$. If $v$ is greater than $high$, we stop searching. If $v$ is in the range $[low, high]$, we add $v$ to the answer. Then, we need to record the last digit of $v$ as $x$. If $x \gt 0$, we add $v \times 10 + x - 1$ to the queue. If $x \lt 9$, we add $v \times 10 + x + 1$ to the queue. Repeat the above steps until the queue is empty.
+
+The time complexity is $O(10 \times 2^{\log M})$, and the space complexity is $O(2^{\log M})$, where $M$ is the number of digits in $high$.
+
 <!-- tabs:start -->
 
 ### **Python3**
@@ -102,17 +110,29 @@ class Solution {
 public:
     vector<int> countSteppingNumbers(int low, int high) {
         vector<int> ans;
-        if (low == 0) ans.push_back(0);
+        if (low == 0) {
+            ans.push_back(0);
+        }
         queue<long long> q;
-        for (int i = 1; i < 10; ++i) q.push(i);
+        for (int i = 1; i < 10; ++i) {
+            q.push(i);
+        }
         while (!q.empty()) {
-            int v = q.front();
+            long long v = q.front();
             q.pop();
-            if (v > high) break;
-            if (v >= low) ans.push_back(v);
+            if (v > high) {
+                break;
+            }
+            if (v >= low) {
+                ans.push_back(v);
+            }
             int x = v % 10;
-            if (x) q.push(1ll * v * 10 + x - 1);
-            if (x < 9) q.push(1ll * v * 10 + x + 1);
+            if (x > 0) {
+                q.push(v * 10 + x - 1);
+            }
+            if (x < 9) {
+                q.push(v * 10 + x + 1);
+            }
         }
         return ans;
     }
@@ -149,6 +169,38 @@ func countSteppingNumbers(low int, high int) []int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function countSteppingNumbers(low: number, high: number): number[] {
+    const ans: number[] = [];
+    if (low === 0) {
+        ans.push(0);
+    }
+    const q: number[] = [];
+    for (let i = 1; i < 10; ++i) {
+        q.push(i);
+    }
+    while (q.length) {
+        const v = q.shift()!;
+        if (v > high) {
+            break;
+        }
+        if (v >= low) {
+            ans.push(v);
+        }
+        const x = v % 10;
+        if (x > 0) {
+            q.push(v * 10 + x - 1);
+        }
+        if (x < 9) {
+            q.push(v * 10 + x + 1);
+        }
+    }
+    return ans;
+}
+```
+
 ### **...**
 
 ```

solution/1200-1299/1215.Stepping Numbers/Solution.cpp

@@ -1,19 +1,31 @@
-class Solution {
-public:
-    vector<int> countSteppingNumbers(int low, int high) {
-        vector<int> ans;
-        if (low == 0) ans.push_back(0);
-        queue<long long> q;
-        for (int i = 1; i < 10; ++i) q.push(i);
-        while (!q.empty()) {
-            int v = q.front();
-            q.pop();
-            if (v > high) break;
-            if (v >= low) ans.push_back(v);
-            int x = v % 10;
-            if (x) q.push(1ll * v * 10 + x - 1);
-            if (x < 9) q.push(1ll * v * 10 + x + 1);
-        }
-        return ans;
-    }
+class Solution {
+public:
+    vector<int> countSteppingNumbers(int low, int high) {
+        vector<int> ans;
+        if (low == 0) {
+            ans.push_back(0);
+        }
+        queue<long long> q;
+        for (int i = 1; i < 10; ++i) {
+            q.push(i);
+        }
+        while (!q.empty()) {
+            long long v = q.front();
+            q.pop();
+            if (v > high) {
+                break;
+            }
+            if (v >= low) {
+                ans.push_back(v);
+            }
+            int x = v % 10;
+            if (x > 0) {
+                q.push(v * 10 + x - 1);
+            }
+            if (x < 9) {
+                q.push(v * 10 + x + 1);
+            }
+        }
+        return ans;
+    }
 };

solution/1200-1299/1215.Stepping Numbers/Solution.ts

@@ -0,0 +1,27 @@
+function countSteppingNumbers(low: number, high: number): number[] {
+    const ans: number[] = [];
+    if (low === 0) {
+        ans.push(0);
+    }
+    const q: number[] = [];
+    for (let i = 1; i < 10; ++i) {
+        q.push(i);
+    }
+    while (q.length) {
+        const v = q.shift()!;
+        if (v > high) {
+            break;
+        }
+        if (v >= low) {
+            ans.push(v);
+        }
+        const x = v % 10;
+        if (x > 0) {
+            q.push(v * 10 + x - 1);
+        }
+        if (x < 9) {
+            q.push(v * 10 + x + 1);
+        }
+    }
+    return ans;
+}

solution/1200-1299/1216.Valid Palindrome III/README.md

@@ -170,6 +170,63 @@ func max(a, b int) int {
 }
 ```
 
+### **TypeScript**
+
+```ts
+function isValidPalindrome(s: string, k: number): boolean {
+    const n = s.length;
+    const f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => 0));
+    for (let i = 0; i < n; ++i) {
+        f[i][i] = 1;
+    }
+    for (let i = n - 2; ~i; --i) {
+        for (let j = i + 1; j < n; ++j) {
+            if (s[i] === s[j]) {
+                f[i][j] = f[i + 1][j - 1] + 2;
+            } else {
+                f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
+            }
+            if (f[i][j] + k >= n) {
+                return true;
+            }
+        }
+    }
+    return false;
+}
+```
+
+### **Rust**
+
+```rust
+impl Solution {
+    pub fn is_valid_palindrome(s: String, k: i32) -> bool {
+        let s = s.as_bytes();
+        let n = s.len();
+        let mut f = vec![vec![0; n]; n];
+
+        for i in 0..n {
+            f[i][i] = 1;
+        }
+
+        for i in (0..n - 2).rev() {
+            for j in (i + 1)..n {
+                if s[i] == s[j] {
+                    f[i][j] = f[i + 1][j - 1] + 2;
+                } else {
+                    f[i][j] = std::cmp::max(f[i + 1][j], f[i][j - 1]);
+                }
+
+                if f[i][j] + k >= n as i32 {
+                    return true;
+                }
+            }
+        }
+
+        false
+    }
+}
+```
+
 ### **...**
 
 ```