diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/README.md b/solution/2100-2199/2119.A Number After a Double Reversal/README.md
index 117583e9746f4..a462fb9e6f900 100644
--- a/solution/2100-2199/2119.A Number After a Double Reversal/README.md
+++ b/solution/2100-2199/2119.A Number After a Double Reversal/README.md
@@ -62,7 +62,11 @@ tags:
<!-- solution:start -->
-### 方法一
+### 方法一：数学
+
+如果数字是 $0$，或者数字的个位不是 $0$，那么反转两次后的数字一定和原数字相等。
+
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
<!-- tabs:start -->
@@ -103,6 +107,36 @@ func isSameAfterReversals(num int) bool {
}
```
+#### TypeScript
+
+```ts
+function isSameAfterReversals(num: number): boolean {
+ return num === 0 || num % 10 !== 0;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+ pub fn is_same_after_reversals(num: i32) -> bool {
+ num == 0 || num % 10 != 0
+ }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number} num
+ * @return {boolean}
+ */
+var isSameAfterReversals = function (num) {
+ return num === 0 || num % 10 !== 0;
+};
+```
+
<!-- tabs:end -->
<!-- solution:end -->
diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/README_EN.md b/solution/2100-2199/2119.A Number After a Double Reversal/README_EN.md
index b8c12ddd1000e..41dc16b2accd3 100644
--- a/solution/2100-2199/2119.A Number After a Double Reversal/README_EN.md
+++ b/solution/2100-2199/2119.A Number After a Double Reversal/README_EN.md
@@ -64,7 +64,11 @@ tags:
<!-- solution:start -->
-### Solution 1
+### Solution 1: Mathematics
+
+If the number is $0$, or the last digit of the number is not $0$, then the number after reversing twice will be the same as the original number.
+
+The time complexity is $O(1)$, and the space complexity is $O(1)$.
<!-- tabs:start -->
@@ -105,6 +109,36 @@ func isSameAfterReversals(num int) bool {
}
```
+#### TypeScript
+
+```ts
+function isSameAfterReversals(num: number): boolean {
+ return num === 0 || num % 10 !== 0;
+}
+```
+
+#### Rust
+
+```rust
+impl Solution {
+ pub fn is_same_after_reversals(num: i32) -> bool {
+ num == 0 || num % 10 != 0
+ }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number} num
+ * @return {boolean}
+ */
+var isSameAfterReversals = function (num) {
+ return num === 0 || num % 10 !== 0;
+};
+```
+
<!-- tabs:end -->
<!-- solution:end -->
diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/Solution.js b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.js
new file mode 100644
index 0000000000000..e971a7af9ec2c
--- /dev/null
+++ b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.js
@@ -0,0 +1,7 @@
+/**
+ * @param {number} num
+ * @return {boolean}
+ */
+var isSameAfterReversals = function (num) {
+ return num === 0 || num % 10 !== 0;
+};
diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/Solution.rs b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.rs
new file mode 100644
index 0000000000000..516887ac60f3f
--- /dev/null
+++ b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.rs
@@ -0,0 +1,5 @@
+impl Solution {
+ pub fn is_same_after_reversals(num: i32) -> bool {
+ num == 0 || num % 10 != 0
+ }
+}
diff --git a/solution/2100-2199/2119.A Number After a Double Reversal/Solution.ts b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.ts
new file mode 100644
index 0000000000000..6c0cd001ad95e
--- /dev/null
+++ b/solution/2100-2199/2119.A Number After a Double Reversal/Solution.ts
@@ -0,0 +1,3 @@
+function isSameAfterReversals(num: number): boolean {
+ return num === 0 || num % 10 !== 0;
+}
diff --git a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md
index 69bf1511aff3b..b5ac84cbc3087 100644
--- a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md
+++ b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README.md
@@ -54,7 +54,7 @@ tags:
### 方法一：数学
-假设三个连续的整数分别为 $x-1$, $x$, $x+1$，则它们的和为 $3x$，因此 $num$ 必须是 $3$ 的倍数。如果 $num$ 不是 $3$ 的倍数，则无法表示成三个连续整数的和，返回空数组。否则，令 $x = \frac{num}{3}$，则 $x-1$, $x$, $x+1$ 就是三个连续整数，它们的和为 $num$。
+我们假设三个连续的整数分别为 $x-1$, $x$, $x+1$，则它们的和为 $3x$，因此 $\textit{num}$ 必须是 $3$ 的倍数。如果 $\textit{num}$ 不是 $3$ 的倍数，则无法表示成三个连续整数的和，返回空数组。否则，令 $x = \frac{\textit{num}}{3}$，则 $x-1$, $x$, $x+1$ 就是三个连续整数，它们的和为 $\textit{num}$。
时间复杂度 $O(1)$，空间复杂度 $O(1)$。
@@ -122,6 +122,36 @@ function sumOfThree(num: number): number[] {
}
```
+#### Rust
+
+```rust
+impl Solution {
+ pub fn sum_of_three(num: i64) -> Vec<i64> {
+ if num % 3 != 0 {
+ return Vec::new();
+ }
+ let x = num / 3;
+ vec![x - 1, x, x + 1]
+ }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number} num
+ * @return {number[]}
+ */
+var sumOfThree = function (num) {
+ if (num % 3) {
+ return [];
+ }
+ const x = Math.floor(num / 3);
+ return [x - 1, x, x + 1];
+};
+```
+
<!-- tabs:end -->
<!-- solution:end -->
diff --git a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md
index e496c2413f7c7..c9c665e1eba37 100644
--- a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md
+++ b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/README_EN.md
@@ -54,7 +54,7 @@ tags:
### Solution 1: Mathematics
-Assume the three consecutive integers are $x-1$, $x$, and $x+1$. Their sum is $3x$, so $num$ must be a multiple of $3$. If $num$ is not a multiple of $3$, it cannot be expressed as the sum of three consecutive integers, and we return an empty array. Otherwise, let $x = \frac{num}{3}$, then $x-1$, $x$, and $x+1$ are the three consecutive integers whose sum is $num$.
+Assume the three consecutive integers are $x-1$, $x$, and $x+1$. Their sum is $3x$, so $\textit{num}$ must be a multiple of $3$. If $\textit{num}$ is not a multiple of $3$, it cannot be represented as the sum of three consecutive integers, and we return an empty array. Otherwise, let $x = \frac{\textit{num}}{3}$, then $x-1$, $x$, and $x+1$ are the three consecutive integers whose sum is $\textit{num}$.
The time complexity is $O(1)$, and the space complexity is $O(1)$.
@@ -122,6 +122,36 @@ function sumOfThree(num: number): number[] {
}
```
+#### Rust
+
+```rust
+impl Solution {
+ pub fn sum_of_three(num: i64) -> Vec<i64> {
+ if num % 3 != 0 {
+ return Vec::new();
+ }
+ let x = num / 3;
+ vec![x - 1, x, x + 1]
+ }
+}
+```
+
+#### JavaScript
+
+```js
+/**
+ * @param {number} num
+ * @return {number[]}
+ */
+var sumOfThree = function (num) {
+ if (num % 3) {
+ return [];
+ }
+ const x = Math.floor(num / 3);
+ return [x - 1, x, x + 1];
+};
+```
+
<!-- tabs:end -->
<!-- solution:end -->
diff --git a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.js b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.js
new file mode 100644
index 0000000000000..548dbf71cc417
--- /dev/null
+++ b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.js
@@ -0,0 +1,11 @@
+/**
+ * @param {number} num
+ * @return {number[]}
+ */
+var sumOfThree = function (num) {
+ if (num % 3) {
+ return [];
+ }
+ const x = Math.floor(num / 3);
+ return [x - 1, x, x + 1];
+};
diff --git a/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.rs b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.rs
new file mode 100644
index 0000000000000..1079c374a223a
--- /dev/null
+++ b/solution/2100-2199/2177.Find Three Consecutive Integers That Sum to a Given Number/Solution.rs
@@ -0,0 +1,9 @@
+impl Solution {
+ pub fn sum_of_three(num: i64) -> Vec<i64> {
+ if num % 3 != 0 {
+ return Vec::new();
+ }
+ let x = num / 3;
+ vec![x - 1, x, x + 1]
+ }
+}
diff --git a/solution/2600-2699/2674.Split a Circular Linked List/README.md b/solution/2600-2699/2674.Split a Circular Linked List/README.md
index 0abbed8f317f7..80957a399b316 100644
--- a/solution/2600-2699/2674.Split a Circular Linked List/README.md
+++ b/solution/2600-2699/2674.Split a Circular Linked List/README.md
@@ -46,7 +46,7 @@ tags:
<p><strong>提示：</strong></p>
<ul>
- <li><code>list</code> 中的节点数范围为 <code>[2, 105]</code></li>
+ <li><code>list</code> 中的节点数范围为 <code>[2, 10<sup>5</sup>]</code></li>
<li><code>0 <= Node.val <= 10<sup>9</sup></code></li>
<li><code>LastNode.next = FirstNode</code> ，其中 <code>LastNode</code> 是链表的最后一个节点，<code>FirstNode</code> 是第一个节点。</li>
</ul>