feat: update solutions (doocs#2258)

Author: yanglbme

solution/0000-0099/0088.Merge Sorted Array/README.md

@@ -131,6 +131,18 @@ function merge(nums1: number[], m: number, nums2: number[], n: number): void {
 }
 ```
 
+```ts
+/**
+ Do not return anything, modify nums1 in-place instead.
+ */
+function merge(nums1: number[], m: number, nums2: number[], n: number): void {
+    nums1.length = m;
+    nums2.length = n;
+    nums1.push(...nums2);
+    nums1.sort((a, b) => a - b);
+}
+```
+
 ```rust
 impl Solution {
     pub fn merge(nums1: &mut Vec<i32>, m: i32, nums2: &mut Vec<i32>, n: i32) {
@@ -198,22 +210,4 @@ class Solution {
 
 <!-- tabs:end -->
 
-### 方法二
-
-<!-- tabs:start -->
-
-```ts
-/**
- Do not return anything, modify nums1 in-place instead.
- */
-function merge(nums1: number[], m: number, nums2: number[], n: number): void {
-    nums1.length = m;
-    nums2.length = n;
-    nums1.push(...nums2);
-    nums1.sort((a, b) => a - b);
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

solution/0000-0099/0088.Merge Sorted Array/README_EN.md

@@ -126,6 +126,18 @@ function merge(nums1: number[], m: number, nums2: number[], n: number): void {
 }
 ```
 
+```ts
+/**
+ Do not return anything, modify nums1 in-place instead.
+ */
+function merge(nums1: number[], m: number, nums2: number[], n: number): void {
+    nums1.length = m;
+    nums2.length = n;
+    nums1.push(...nums2);
+    nums1.sort((a, b) => a - b);
+}
+```
+
 ```rust
 impl Solution {
     pub fn merge(nums1: &mut Vec<i32>, m: i32, nums2: &mut Vec<i32>, n: i32) {
@@ -193,22 +205,4 @@ class Solution {
 
 <!-- tabs:end -->
 
-### Solution 2
-
-<!-- tabs:start -->
-
-```ts
-/**
- Do not return anything, modify nums1 in-place instead.
- */
-function merge(nums1: number[], m: number, nums2: number[], n: number): void {
-    nums1.length = m;
-    nums2.length = n;
-    nums1.push(...nums2);
-    nums1.sort((a, b) => a - b);
-}
-```
-
-<!-- tabs:end -->
-
 <!-- end -->

solution/0000-0099/0091.Decode Ways/README.md

@@ -75,8 +75,6 @@
 
 时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是字符串的长度。
 
-我们注意到，状态 $f[i]$ 仅与状态 $f[i-1]$ 和状态 $f[i-2]$ 有关，而与其他状态无关，因此我们可以使用两个变量代替这两个状态，使得原来的空间复杂度 $O(n)$ 降低至 $O(1)$。
-
 <!-- tabs:start -->
 
 ```python
@@ -187,7 +185,7 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### 方法二
+我们注意到，状态 $f[i]$ 仅与状态 $f[i-1]$ 和状态 $f[i-2]$ 有关，而与其他状态无关，因此我们可以使用两个变量代替这两个状态，使得原来的空间复杂度 $O(n)$ 降低至 $O(1)$。
 
 <!-- tabs:start -->
 

solution/0000-0099/0091.Decode Ways/README_EN.md

@@ -182,7 +182,7 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### Solution 2
+We notice that the state $f[i]$ is only related to the states $f[i-1]$ and $f[i-2]$, and is irrelevant to other states. Therefore, we can use two variables to replace these two states, reducing the original space complexity from $O(n)$ to $O(1)$.
 
 <!-- tabs:start -->
 

solution/0000-0099/0097.Interleaving String/README.md

@@ -357,8 +357,6 @@ $$
 
 时间复杂度 $O(m \times n)$，空间复杂度 $O(m \times n)$。其中 $m$ 和 $n$ 分别是字符串 $s_1$ 和 $s_2$ 的长度。
 
-我们注意到，状态 $f[i][j]$ 只和状态 $f[i - 1][j]$、$f[i][j - 1]$、$f[i - 1][j - 1]$ 有关，因此我们可以使用滚动数组优化空间复杂度，将空间复杂度优化到 $O(n)$。
-
 <!-- tabs:start -->
 
 ```python
@@ -508,7 +506,7 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### 方法三
+我们注意到，状态 $f[i][j]$ 只和状态 $f[i - 1][j]$、$f[i][j - 1]$、$f[i - 1][j - 1]$ 有关，因此我们可以使用滚动数组优化空间复杂度，将空间复杂度优化到 $O(n)$。
 
 <!-- tabs:start -->
 

solution/0000-0099/0097.Interleaving String/README_EN.md

@@ -357,8 +357,6 @@ The answer is $f[m][n]$.
 
 The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the lengths of strings $s_1$ and $s_2$ respectively.
 
-We notice that the state $f[i][j]$ is only related to the states $f[i - 1][j]$, $f[i][j - 1]$, and $f[i - 1][j - 1]$. Therefore, we can use a rolling array to optimize the space complexity, reducing the original space complexity from $O(m \times n)$ to $O(n)$.
-
 <!-- tabs:start -->
 
 ```python
@@ -508,7 +506,7 @@ public class Solution {
 
 <!-- tabs:end -->
 
-### Solution 3
+We notice that the state $f[i][j]$ is only related to the states $f[i - 1][j]$, $f[i][j - 1]$, and $f[i - 1][j - 1]$. Therefore, we can use a rolling array to optimize the space complexity, reducing the original space complexity from $O(m \times n)$ to $O(n)$.
 
 <!-- tabs:start -->