feat: add solutions to lc problems: No.2119,2177

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 -->

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 -->

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;
+};

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
+    }
+}

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;
+}

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 -->

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 -->

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];
+};

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]
+    }
+}

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 &lt;= Node.val &lt;= 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>