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feat: add solutions to lc problems: No.1215,1216 (doocs#1875)
* No.1215.Stepping Numbers * No.1216.Valid Palindrome III
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Diff for: ‎solution/1200-1299/1215.Stepping Numbers/README.md

+57-7
Original file line numberDiff line numberDiff line change
@@ -34,6 +34,12 @@
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**方法一：BFS**
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首先，如果 $low$ 为 $0$，那么我们需要将 $0$ 加入答案中。
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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。
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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。
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<!-- tabs:start -->
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### **Python3**
@@ -104,17 +110,29 @@ class Solution {
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public:
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vector<int> countSteppingNumbers(int low, int high) {
106112
vector<int> ans;
107-
if (low == 0) ans.push_back(0);
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if (low == 0) {
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ans.push_back(0);
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}
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queue<long long> q;
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for (int i = 1; i < 10; ++i) q.push(i);
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for (int i = 1; i < 10; ++i) {
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q.push(i);
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}
110120
while (!q.empty()) {
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int v = q.front();
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long long v = q.front();
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q.pop();
113-
if (v > high) break;
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if (v >= low) ans.push_back(v);
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if (v > high) {
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break;
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}
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if (v >= low) {
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ans.push_back(v);
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}
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int x = v % 10;
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if (x) q.push(1ll * v * 10 + x - 1);
117-
if (x < 9) q.push(1ll * v * 10 + x + 1);
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if (x > 0) {
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q.push(v * 10 + x - 1);
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}
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if (x < 9) {
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q.push(v * 10 + x + 1);
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}
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}
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return ans;
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}
@@ -151,6 +169,38 @@ func countSteppingNumbers(low int, high int) []int {
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}
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```
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### **TypeScript**
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```ts
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function countSteppingNumbers(low: number, high: number): number[] {
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const ans: number[] = [];
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if (low === 0) {
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ans.push(0);
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}
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const q: number[] = [];
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for (let i = 1; i < 10; ++i) {
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q.push(i);
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}
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while (q.length) {
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const v = q.shift()!;
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if (v > high) {
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break;
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}
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if (v >= low) {
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ans.push(v);
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}
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const x = v % 10;
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if (x > 0) {
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q.push(v * 10 + x - 1);
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}
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if (x < 9) {
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q.push(v * 10 + x + 1);
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}
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}
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return ans;
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}
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```
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### **...**
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```

Diff for: ‎solution/1200-1299/1215.Stepping Numbers/README_EN.md

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Original file line numberDiff line numberDiff line change
@@ -36,6 +36,14 @@
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## Solutions
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**Solution 1: BFS**
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First, if $low$ is $0$, we need to add $0$ to the answer.
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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.
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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 $high$.
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<!-- tabs:start -->
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4149
### **Python3**
@@ -102,17 +110,29 @@ class Solution {
102110
public:
103111
vector<int> countSteppingNumbers(int low, int high) {
104112
vector<int> ans;
105-
if (low == 0) ans.push_back(0);
113+
if (low == 0) {
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ans.push_back(0);
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}
106116
queue<long long> q;
107-
for (int i = 1; i < 10; ++i) q.push(i);
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for (int i = 1; i < 10; ++i) {
118+
q.push(i);
119+
}
108120
while (!q.empty()) {
109-
int v = q.front();
121+
long long v = q.front();
110122
q.pop();
111-
if (v > high) break;
112-
if (v >= low) ans.push_back(v);
123+
if (v > high) {
124+
break;
125+
}
126+
if (v >= low) {
127+
ans.push_back(v);
128+
}
113129
int x = v % 10;
114-
if (x) q.push(1ll * v * 10 + x - 1);
115-
if (x < 9) q.push(1ll * v * 10 + x + 1);
130+
if (x > 0) {
131+
q.push(v * 10 + x - 1);
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}
133+
if (x < 9) {
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q.push(v * 10 + x + 1);
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}
116136
}
117137
return ans;
118138
}
@@ -149,6 +169,38 @@ func countSteppingNumbers(low int, high int) []int {
149169
}
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```
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### **TypeScript**
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```ts
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function countSteppingNumbers(low: number, high: number): number[] {
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const ans: number[] = [];
177+
if (low === 0) {
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ans.push(0);
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}
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const q: number[] = [];
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for (let i = 1; i < 10; ++i) {
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q.push(i);
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}
184+
while (q.length) {
185+
const v = q.shift()!;
186+
if (v > high) {
187+
break;
188+
}
189+
if (v >= low) {
190+
ans.push(v);
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}
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const x = v % 10;
193+
if (x > 0) {
194+
q.push(v * 10 + x - 1);
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}
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if (x < 9) {
197+
q.push(v * 10 + x + 1);
198+
}
199+
}
200+
return ans;
201+
}
202+
```
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### **...**
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```

Diff for: ‎solution/1200-1299/1215.Stepping Numbers/Solution.cpp

+30-18
Original file line numberDiff line numberDiff line change
@@ -1,19 +1,31 @@
1-
class Solution {
2-
public:
3-
vector<int> countSteppingNumbers(int low, int high) {
4-
vector<int> ans;
5-
if (low == 0) ans.push_back(0);
6-
queue<long long> q;
7-
for (int i = 1; i < 10; ++i) q.push(i);
8-
while (!q.empty()) {
9-
int v = q.front();
10-
q.pop();
11-
if (v > high) break;
12-
if (v >= low) ans.push_back(v);
13-
int x = v % 10;
14-
if (x) q.push(1ll * v * 10 + x - 1);
15-
if (x < 9) q.push(1ll * v * 10 + x + 1);
16-
}
17-
return ans;
18-
}
1+
class Solution {
2+
public:
3+
vector<int> countSteppingNumbers(int low, int high) {
4+
vector<int> ans;
5+
if (low == 0) {
6+
ans.push_back(0);
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}
8+
queue<long long> q;
9+
for (int i = 1; i < 10; ++i) {
10+
q.push(i);
11+
}
12+
while (!q.empty()) {
13+
long long v = q.front();
14+
q.pop();
15+
if (v > high) {
16+
break;
17+
}
18+
if (v >= low) {
19+
ans.push_back(v);
20+
}
21+
int x = v % 10;
22+
if (x > 0) {
23+
q.push(v * 10 + x - 1);
24+
}
25+
if (x < 9) {
26+
q.push(v * 10 + x + 1);
27+
}
28+
}
29+
return ans;
30+
}
1931
};

Diff for: ‎solution/1200-1299/1215.Stepping Numbers/Solution.ts

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Original file line numberDiff line numberDiff line change
@@ -0,0 +1,27 @@
1+
function countSteppingNumbers(low: number, high: number): number[] {
2+
const ans: number[] = [];
3+
if (low === 0) {
4+
ans.push(0);
5+
}
6+
const q: number[] = [];
7+
for (let i = 1; i < 10; ++i) {
8+
q.push(i);
9+
}
10+
while (q.length) {
11+
const v = q.shift()!;
12+
if (v > high) {
13+
break;
14+
}
15+
if (v >= low) {
16+
ans.push(v);
17+
}
18+
const x = v % 10;
19+
if (x > 0) {
20+
q.push(v * 10 + x - 1);
21+
}
22+
if (x < 9) {
23+
q.push(v * 10 + x + 1);
24+
}
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}
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return ans;
27+
}

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

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Original file line numberDiff line numberDiff line change
@@ -170,6 +170,63 @@ func max(a, b int) int {
170170
}
171171
```
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### **TypeScript**
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```ts
176+
function isValidPalindrome(s: string, k: number): boolean {
177+
const n = s.length;
178+
const f: number[][] = Array.from({ length: n }, () => Array.from({ length: n }, () => 0));
179+
for (let i = 0; i < n; ++i) {
180+
f[i][i] = 1;
181+
}
182+
for (let i = n - 2; ~i; --i) {
183+
for (let j = i + 1; j < n; ++j) {
184+
if (s[i] === s[j]) {
185+
f[i][j] = f[i + 1][j - 1] + 2;
186+
} else {
187+
f[i][j] = Math.max(f[i + 1][j], f[i][j - 1]);
188+
}
189+
if (f[i][j] + k >= n) {
190+
return true;
191+
}
192+
}
193+
}
194+
return false;
195+
}
196+
```
197+
198+
### **Rust**
199+
200+
```rust
201+
impl Solution {
202+
pub fn is_valid_palindrome(s: String, k: i32) -> bool {
203+
let s = s.as_bytes();
204+
let n = s.len();
205+
let mut f = vec![vec![0; n]; n];
206+
207+
for i in 0..n {
208+
f[i][i] = 1;
209+
}
210+
211+
for i in (0..n - 2).rev() {
212+
for j in (i + 1)..n {
213+
if s[i] == s[j] {
214+
f[i][j] = f[i + 1][j - 1] + 2;
215+
} else {
216+
f[i][j] = std::cmp::max(f[i + 1][j], f[i][j - 1]);
217+
}
218+
219+
if f[i][j] + k >= n as i32 {
220+
return true;
221+
}
222+
}
223+
}
224+
225+
false
226+
}
227+
}
228+
```
229+
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### **...**
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```

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