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feat: add solutions to lc problems #2601

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Apr 16, 2024
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feat: add solutions to lc problems #2601

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feat: add solutions to lc problems
  • Loading branch information
@yanglbme
yanglbme committed Apr 16, 2024
commit a3b1d66ee1bdfaf00a63b71a317e3ad1fc81caa7
6 changes: 3 additions & 3 deletions solution/0900-0999/0908.Smallest Range I/README.md
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Expand Up @@ -104,9 +104,9 @@ func smallestRangeI(nums []int, k int) int {

```ts
function smallestRangeI(nums: number[], k: number): number {
const max = nums.reduce((r, v) => Math.max(r, v));
const min = nums.reduce((r, v) => Math.min(r, v));
return Math.max(max - min - k * 2, 0);
const mx = Math.max(...nums);
const mi = Math.min(...nums);
return Math.max(mx - mi - k * 2, 0);
}
```

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6 changes: 3 additions & 3 deletions solution/0900-0999/0908.Smallest Range I/README_EN.md
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Expand Up @@ -100,9 +100,9 @@ func smallestRangeI(nums []int, k int) int {

```ts
function smallestRangeI(nums: number[], k: number): number {
const max = nums.reduce((r, v) => Math.max(r, v));
const min = nums.reduce((r, v) => Math.min(r, v));
return Math.max(max - min - k * 2, 0);
const mx = Math.max(...nums);
const mi = Math.min(...nums);
return Math.max(mx - mi - k * 2, 0);
}
```

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6 changes: 3 additions & 3 deletions solution/0900-0999/0908.Smallest Range I/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,5 +1,5 @@
function smallestRangeI(nums: number[], k: number): number {
const max = nums.reduce((r, v) => Math.max(r, v));
const min = nums.reduce((r, v) => Math.min(r, v));
return Math.max(max - min - k * 2, 0);
const mx = Math.max(...nums);
const mi = Math.min(...nums);
return Math.max(mx - mi - k * 2, 0);
}
13 changes: 13 additions & 0 deletions solution/0900-0999/0910.Smallest Range II/README.md
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Expand Up @@ -126,6 +126,19 @@ func smallestRangeII(nums []int, k int) int {
}
```

```ts
function smallestRangeII(nums: number[], k: number): number {
nums.sort((a, b) => a - b);
let ans = nums.at(-1)! - nums[0];
for (let i = 1; i < nums.length; ++i) {
const mi = Math.min(nums[0] + k, nums[i] - k);
const mx = Math.max(nums.at(-1)! - k, nums[i - 1] + k);
ans = Math.min(ans, mx - mi);
}
return ans;
}
```

<!-- tabs:end -->

<!-- end -->
13 changes: 13 additions & 0 deletions solution/0900-0999/0910.Smallest Range II/README_EN.md
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Expand Up @@ -119,6 +119,19 @@ func smallestRangeII(nums []int, k int) int {
}
```

```ts
function smallestRangeII(nums: number[], k: number): number {
nums.sort((a, b) => a - b);
let ans = nums.at(-1)! - nums[0];
for (let i = 1; i < nums.length; ++i) {
const mi = Math.min(nums[0] + k, nums[i] - k);
const mx = Math.max(nums.at(-1)! - k, nums[i - 1] + k);
ans = Math.min(ans, mx - mi);
}
return ans;
}
```

<!-- tabs:end -->

<!-- end -->
10 changes: 10 additions & 0 deletions solution/0900-0999/0910.Smallest Range II/Solution.ts
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@@ -0,0 +1,10 @@
function smallestRangeII(nums: number[], k: number): number {
nums.sort((a, b) => a - b);
let ans = nums.at(-1)! - nums[0];
for (let i = 1; i < nums.length; ++i) {
const mi = Math.min(nums[0] + k, nums[i] - k);
const mx = Math.max(nums.at(-1)! - k, nums[i - 1] + k);
ans = Math.min(ans, mx - mi);
}
return ans;
}
72 changes: 40 additions & 32 deletions solution/0900-0999/0917.Reverse Only Letters/README.md
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Expand Up @@ -61,47 +61,51 @@

## 解法

### 方法一
### 方法一：双指针

我们用两个指针 $i$ 和 $j$ 分别指向字符串的头部和尾部。当 $i < j$ 时，我们不断地移动 $i$ 和 $j$，直到 $i$ 指向一个英文字母，并且 $j$ 指向一个英文字母，然后交换 $s[i]$ 和 $s[j]$。最后返回字符串即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为字符串的长度。

<!-- tabs:start -->

```python
class Solution:
def reverseOnlyLetters(self, s: str) -> str:
s = list(s)
i, j = 0, len(s) - 1
cs = list(s)
i, j = 0, len(cs) - 1
while i < j:
while i < j and not s[i].isalpha():
while i < j and not cs[i].isalpha():
i += 1
while i < j and not s[j].isalpha():
while i < j and not cs[j].isalpha():
j -= 1
if i < j:
s[i], s[j] = s[j], s[i]
cs[i], cs[j] = cs[j], cs[i]
i, j = i + 1, j - 1
return ''.join(s)
return "".join(cs)
```

```java
class Solution {
public String reverseOnlyLetters(String s) {
char[] chars = s.toCharArray();
int i = 0, j = s.length() - 1;
char[] cs = s.toCharArray();
int i = 0, j = cs.length - 1;
while (i < j) {
while (i < j && !Character.isLetter(chars[i])) {
while (i < j && !Character.isLetter(cs[i])) {
++i;
}
while (i < j && !Character.isLetter(chars[j])) {
while (i < j && !Character.isLetter(cs[j])) {
--j;
}
if (i < j) {
char t = chars[i];
chars[i] = chars[j];
chars[j] = t;
char t = cs[i];
cs[i] = cs[j];
cs[j] = t;
++i;
--j;
}
}
return new String(chars);
return new String(cs);
}
}
```
Expand All @@ -112,13 +116,15 @@ public:
string reverseOnlyLetters(string s) {
int i = 0, j = s.size() - 1;
while (i < j) {
while (i < j && !isalpha(s[i])) ++i;
while (i < j && !isalpha(s[j])) --j;
if (i < j) {
swap(s[i], s[j]);
while (i < j && !isalpha(s[i])) {
++i;
}
while (i < j && !isalpha(s[j])) {
--j;
}
if (i < j) {
swap(s[i++], s[j--]);
}
}
return s;
}
Expand All @@ -127,39 +133,41 @@ public:

```go
func reverseOnlyLetters(s string) string {
ans := []rune(s)
cs := []rune(s)
i, j := 0, len(s)-1
for i < j {
for i < j && !unicode.IsLetter(ans[i]) {
for i < j && !unicode.IsLetter(cs[i]) {
i++
}
for i < j && !unicode.IsLetter(ans[j]) {
for i < j && !unicode.IsLetter(cs[j]) {
j--
}
if i < j {
ans[i], ans[j] = ans[j], ans[i]
cs[i], cs[j] = cs[j], cs[i]
i++
j--
}
}
return string(ans)
return string(cs)
}
```

```ts
function reverseOnlyLetters(s: string): string {
const n = s.length;
let i = 0,
j = n - 1;
let ans = [...s];
const cs = [...s];
let [i, j] = [0, cs.length - 1];
while (i < j) {
while (!/[a-zA-Z]/.test(ans[i]) && i < j) i++;
while (!/[a-zA-Z]/.test(ans[j]) && i < j) j--;
[ans[i], ans[j]] = [ans[j], ans[i]];
while (!/[a-zA-Z]/.test(cs[i]) && i < j) {
i++;
}
while (!/[a-zA-Z]/.test(cs[j]) && i < j) {
j--;
}
[cs[i], cs[j]] = [cs[j], cs[i]];
i++;
j--;
}
return ans.join('');
return cs.join('');
}
```

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72 changes: 40 additions & 32 deletions solution/0900-0999/0917.Reverse Only Letters/README_EN.md
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Expand Up @@ -37,47 +37,51 @@

## Solutions

### Solution 1
### Solution 1: Two Pointers

We use two pointers $i$ and $j$ to point to the head and tail of the string respectively. When $i < j$, we continuously move $i$ and $j$ until $i$ points to an English letter and $j$ points to an English letter, then we swap $s[i]$ and $s[j]$. Finally, we return the string.

The time complexity is $O(n)$, and the space complexity is $O(n)$, where $n$ is the length of the string.

<!-- tabs:start -->

```python
class Solution:
def reverseOnlyLetters(self, s: str) -> str:
s = list(s)
i, j = 0, len(s) - 1
cs = list(s)
i, j = 0, len(cs) - 1
while i < j:
while i < j and not s[i].isalpha():
while i < j and not cs[i].isalpha():
i += 1
while i < j and not s[j].isalpha():
while i < j and not cs[j].isalpha():
j -= 1
if i < j:
s[i], s[j] = s[j], s[i]
cs[i], cs[j] = cs[j], cs[i]
i, j = i + 1, j - 1
return ''.join(s)
return "".join(cs)
```

```java
class Solution {
public String reverseOnlyLetters(String s) {
char[] chars = s.toCharArray();
int i = 0, j = s.length() - 1;
char[] cs = s.toCharArray();
int i = 0, j = cs.length - 1;
while (i < j) {
while (i < j && !Character.isLetter(chars[i])) {
while (i < j && !Character.isLetter(cs[i])) {
++i;
}
while (i < j && !Character.isLetter(chars[j])) {
while (i < j && !Character.isLetter(cs[j])) {
--j;
}
if (i < j) {
char t = chars[i];
chars[i] = chars[j];
chars[j] = t;
char t = cs[i];
cs[i] = cs[j];
cs[j] = t;
++i;
--j;
}
}
return new String(chars);
return new String(cs);
}
}
```
Expand All @@ -88,13 +92,15 @@ public:
string reverseOnlyLetters(string s) {
int i = 0, j = s.size() - 1;
while (i < j) {
while (i < j && !isalpha(s[i])) ++i;
while (i < j && !isalpha(s[j])) --j;
if (i < j) {
swap(s[i], s[j]);
while (i < j && !isalpha(s[i])) {
++i;
}
while (i < j && !isalpha(s[j])) {
--j;
}
if (i < j) {
swap(s[i++], s[j--]);
}
}
return s;
}
Expand All @@ -103,39 +109,41 @@ public:

```go
func reverseOnlyLetters(s string) string {
ans := []rune(s)
cs := []rune(s)
i, j := 0, len(s)-1
for i < j {
for i < j && !unicode.IsLetter(ans[i]) {
for i < j && !unicode.IsLetter(cs[i]) {
i++
}
for i < j && !unicode.IsLetter(ans[j]) {
for i < j && !unicode.IsLetter(cs[j]) {
j--
}
if i < j {
ans[i], ans[j] = ans[j], ans[i]
cs[i], cs[j] = cs[j], cs[i]
i++
j--
}
}
return string(ans)
return string(cs)
}
```

```ts
function reverseOnlyLetters(s: string): string {
const n = s.length;
let i = 0,
j = n - 1;
let ans = [...s];
const cs = [...s];
let [i, j] = [0, cs.length - 1];
while (i < j) {
while (!/[a-zA-Z]/.test(ans[i]) && i < j) i++;
while (!/[a-zA-Z]/.test(ans[j]) && i < j) j--;
[ans[i], ans[j]] = [ans[j], ans[i]];
while (!/[a-zA-Z]/.test(cs[i]) && i < j) {
i++;
}
while (!/[a-zA-Z]/.test(cs[j]) && i < j) {
j--;
}
[cs[i], cs[j]] = [cs[j], cs[i]];
i++;
j--;
}
return ans.join('');
return cs.join('');
}
```

Expand Down
10 changes: 6 additions & 4 deletions solution/0900-0999/0917.Reverse Only Letters/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -3,13 +3,15 @@ class Solution {
string reverseOnlyLetters(string s) {
int i = 0, j = s.size() - 1;
while (i < j) {
while (i < j && !isalpha(s[i])) ++i;
while (i < j && !isalpha(s[j])) --j;
if (i < j) {
swap(s[i], s[j]);
while (i < j && !isalpha(s[i])) {
++i;
}
while (i < j && !isalpha(s[j])) {
--j;
}
if (i < j) {
swap(s[i++], s[j--]);
}
}
return s;
}
Expand Down
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