feat: update solutions to lc problems: No.344,345 (#2987)

* No.0344.Reverse String
* No.0345.Reverse Vowels of a String

Author: yanglbme

solution/0300-0399/0344.Reverse String/README.md

@@ -154,33 +154,4 @@ var reverseString = function (s) {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def reverseString(self, s: List[str]) -> None:
-        s[:] = s[::-1]
-```
-
-#### TypeScript
-
-```ts
-/**
- Do not return anything, modify s in-place instead.
- */
-function reverseString(s: string[]): void {
-    s.reverse();
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/0300-0399/0344.Reverse String/README_EN.md

@@ -43,7 +43,11 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Two Pointers
+
+We use two pointers $i$ and $j$, initially pointing to the start and end of the array respectively. Each time, we swap the elements at $i$ and $j$, then move $i$ forward and $j$ backward, until $i$ and $j$ meet.
+
+The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.
 
 <!-- tabs:start -->
 
@@ -142,33 +146,4 @@ var reverseString = function (s) {
 
 <!-- solution:end -->
 
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
-    def reverseString(self, s: List[str]) -> None:
-        s[:] = s[::-1]
-```
-
-#### TypeScript
-
-```ts
-/**
- Do not return anything, modify s in-place instead.
- */
-function reverseString(s: string[]): void {
-    s.reverse();
-}
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
 <!-- problem:end -->

solution/0300-0399/0344.Reverse String/Solution2.py

@@ -43,7 +43,13 @@ tags:
 
 <!-- solution:start -->
 
-### Solution 1
+### Solution 1: Two Pointers
+
+We can use two pointers $i$ and $j$, initially pointing to the start and end of the string respectively.
+
+In each loop, we check whether the character at $i$ is a vowel. If it's not, we move $i$ forward. Similarly, we check whether the character at $j$ is a vowel. If it's not, we move $j$ backward. If $i < j$ at this point, then both characters at $i$ and $j$ are vowels, so we swap these two characters. Then, we move $i$ forward and $j$ backward. We continue the above operations until $i \ge j$.
+
+The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(|\Sigma|)$, where $\Sigma$ is the size of the character set.
 
 <!-- tabs:start -->