diff --git a/solution/1900-1999/1929.Concatenation of Array/README.md b/solution/1900-1999/1929.Concatenation of Array/README.md
index cb5bfa919efac..c378e6909f0ca 100644
--- a/solution/1900-1999/1929.Concatenation of Array/README.md
+++ b/solution/1900-1999/1929.Concatenation of Array/README.md
@@ -67,7 +67,11 @@ tags:
<!-- solution:start -->
-### 方法一
+### 方法一：模拟
+
+我们直接根据题目描述模拟，将 $\textit{nums}$ 中的元素依次添加到答案数组中，然后再将 $\textit{nums}$ 中的元素再次添加到答案数组中。
+
+时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
<!-- tabs:start -->
@@ -142,9 +146,7 @@ impl Solution {
* @return {number[]}
*/
var getConcatenation = function (nums) {
- let ans = nums.slice();
- ans.splice(nums.length, 0, ...nums);
- return ans;
+ return [...nums, ...nums];
};
```
diff --git a/solution/1900-1999/1929.Concatenation of Array/README_EN.md b/solution/1900-1999/1929.Concatenation of Array/README_EN.md
index bc384edba3963..26c0427d37fca 100644
--- a/solution/1900-1999/1929.Concatenation of Array/README_EN.md
+++ b/solution/1900-1999/1929.Concatenation of Array/README_EN.md
@@ -60,7 +60,11 @@ tags:
<!-- solution:start -->
-### Solution 1
+### Solution 1: Simulation
+
+We directly simulate according to the problem description by adding the elements of $\textit{nums}$ to the answer array one by one, and then adding the elements of $\textit{nums}$ to the answer array again.
+
+The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
<!-- tabs:start -->
@@ -135,9 +139,7 @@ impl Solution {
* @return {number[]}
*/
var getConcatenation = function (nums) {
- let ans = nums.slice();
- ans.splice(nums.length, 0, ...nums);
- return ans;
+ return [...nums, ...nums];
};
```
diff --git a/solution/1900-1999/1929.Concatenation of Array/Solution.js b/solution/1900-1999/1929.Concatenation of Array/Solution.js
index 86c0ff6a81ef2..a920cb19b9d40 100644
--- a/solution/1900-1999/1929.Concatenation of Array/Solution.js
+++ b/solution/1900-1999/1929.Concatenation of Array/Solution.js
@@ -3,7 +3,5 @@
* @return {number[]}
*/
var getConcatenation = function (nums) {
- let ans = nums.slice();
- ans.splice(nums.length, 0, ...nums);
- return ans;
+ return [...nums, ...nums];
};
diff --git a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md
index f49d208cd3ddb..89286d45008fb 100644
--- a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md
+++ b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README.md
@@ -82,18 +82,14 @@ tags:
```python
class Solution:
def isDecomposable(self, s: str) -> bool:
- i, n = 0, len(s)
cnt2 = 0
- while i < n:
- j = i
- while j < n and s[j] == s[i]:
- j += 1
- if (j - i) % 3 == 1:
+ for _, g in groupby(s):
+ m = len(list(g))
+ if m % 3 == 1:
return False
- cnt2 += (j - i) % 3 == 2
+ cnt2 += m % 3 == 2
if cnt2 > 1:
return False
- i = j
return cnt2 == 1
```
@@ -201,30 +197,4 @@ function isDecomposable(s: string): boolean {
<!-- solution:end -->
-<!-- solution:start -->
-
-### 方法二
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
- def isDecomposable(self, s: str) -> bool:
- cnt2 = 0
- for _, g in groupby(s):
- m = len(list(g))
- if m % 3 == 1:
- return False
- cnt2 += m % 3 == 2
- if cnt2 > 1:
- return False
- return cnt2 == 1
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
<!-- problem:end -->
diff --git a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md
index 10136446eb960..889d99fa3ac1b 100644
--- a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md
+++ b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/README_EN.md
@@ -83,18 +83,14 @@ The time complexity is $O(n)$, where $n$ is the length of the string $s$. The sp
```python
class Solution:
def isDecomposable(self, s: str) -> bool:
- i, n = 0, len(s)
cnt2 = 0
- while i < n:
- j = i
- while j < n and s[j] == s[i]:
- j += 1
- if (j - i) % 3 == 1:
+ for _, g in groupby(s):
+ m = len(list(g))
+ if m % 3 == 1:
return False
- cnt2 += (j - i) % 3 == 2
+ cnt2 += m % 3 == 2
if cnt2 > 1:
return False
- i = j
return cnt2 == 1
```
@@ -202,30 +198,4 @@ function isDecomposable(s: string): boolean {
<!-- solution:end -->
-<!-- solution:start -->
-
-### Solution 2
-
-<!-- tabs:start -->
-
-#### Python3
-
-```python
-class Solution:
- def isDecomposable(self, s: str) -> bool:
- cnt2 = 0
- for _, g in groupby(s):
- m = len(list(g))
- if m % 3 == 1:
- return False
- cnt2 += m % 3 == 2
- if cnt2 > 1:
- return False
- return cnt2 == 1
-```
-
-<!-- tabs:end -->
-
-<!-- solution:end -->
-
<!-- problem:end -->
diff --git a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution2.py b/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution2.py
deleted file mode 100644
index 5b938278d2632..0000000000000
--- a/solution/1900-1999/1933.Check if String Is Decomposable Into Value-Equal Substrings/Solution2.py
+++ /dev/null
@@ -1,11 +0,0 @@
-class Solution:
- def isDecomposable(self, s: str) -> bool:
- cnt2 = 0
- for _, g in groupby(s):
- m = len(list(g))
- if m % 3 == 1:
- return False
- cnt2 += m % 3 == 2
- if cnt2 > 1:
- return False
- return cnt2 == 1