feat: add solutions to lc problems: No.2634,2635

* No.2634.Filter Elements from Array
* No.2635.Apply Transform Over Each Element in Array

Author: yanglbme

solution/2600-2699/2634.Filter Elements from Array/README.md

@@ -55,14 +55,28 @@ const newArray = filter(arr, fn); // [20, 30]
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：遍历**
+
+我们遍历数组 $arr$，对于每个元素 $arr[i]$，如果 $fn(arr[i], i)$ 为真，则将其加入答案数组中。最后返回答案数组即可。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组 $arr$ 的长度。忽略答案的空间消耗，空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **TypeScript**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```ts
-
+function filter(arr: number[], fn: (n: number, i: number) => any): number[] {
+    const ans: number[] = [];
+    for (let i = 0; i < arr.length; ++i) {
+        if (fn(arr[i], i)) {
+            ans.push(arr[i]);
+        }
+    }
+    return ans;
+}
 ```
 
 ### **...**

solution/2600-2699/2634.Filter Elements from Array/README_EN.md

@@ -49,12 +49,26 @@ Falsey values such as 0 should be filtered out
 
 ## Solutions
 
+**Approach 1: Traversal**
+
+We traverse the array $arr$ and for each element $arr[i]$, if $fn(arr[i], i)$ is true, we add it to the answer array. Finally, we return the answer array.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $arr$. Ignoring the space consumption of the answer, the space complexity is $O(1)$.
+
 <!-- tabs:start -->
 
 ### **TypeScript**
 
 ```ts
-
+function filter(arr: number[], fn: (n: number, i: number) => any): number[] {
+    const ans: number[] = [];
+    for (let i = 0; i < arr.length; ++i) {
+        if (fn(arr[i], i)) {
+            ans.push(arr[i]);
+        }
+    }
+    return ans;
+}
 ```
 
 ### **...**

solution/2600-2699/2634.Filter Elements from Array/Solution.ts

@@ -0,0 +1,9 @@
+function filter(arr: number[], fn: (n: number, i: number) => any): number[] {
+    const ans: number[] = [];
+    for (let i = 0; i < arr.length; ++i) {
+        if (fn(arr[i], i)) {
+            ans.push(arr[i]);
+        }
+    }
+    return ans;
+}

solution/2600-2699/2635.Apply Transform Over Each Element in Array/README.md

@@ -55,14 +55,25 @@ const newArray = map(arr, plusone); // [2,3,4]
 
 <!-- 这里可写通用的实现逻辑 -->
 
+**方法一：遍历**
+
+我们遍历数组 $arr$，对于每个元素 $arr[i]$，将其替换为 $fn(arr[i], i)$。最后返回数组 $arr$ 即可。
+
+时间复杂度 $O(n)$，其中 $n$ 为数组 $arr$ 的长度。空间复杂度 $O(1)$。
+
 <!-- tabs:start -->
 
 ### **TypeScript**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```ts
-
+function map(arr: number[], fn: (n: number, i: number) => number): number[] {
+    for (let i = 0; i < arr.length; ++i) {
+        arr[i] = fn(arr[i], i);
+    }
+    return arr;
+}
 ```
 
 ### **...**

solution/2600-2699/2635.Apply Transform Over Each Element in Array/README_EN.md

@@ -48,12 +48,23 @@ The function increases each value in the array by one.
 
 ## Solutions
 
+**Approach 1: traversal**
+
+We traverse the array $arr$, for each element $arr[i]$, replace it with $fn(arr[i], i)$. Finally, return the array $arr$.
+
+The time complexity is $O(n)$, where $n$ is the length of the array $arr$. The space complexity is $O(1)$.
+
 <!-- tabs:start -->
 
 ### **TypeScript**
 
 ```ts
-
+function map(arr: number[], fn: (n: number, i: number) => number): number[] {
+    for (let i = 0; i < arr.length; ++i) {
+        arr[i] = fn(arr[i], i);
+    }
+    return arr;
+}
 ```
 
 ### **...**

solution/2600-2699/2635.Apply Transform Over Each Element in Array/Solution.ts

@@ -0,0 +1,6 @@
+function map(arr: number[], fn: (n: number, i: number) => number): number[] {
+    for (let i = 0; i < arr.length; ++i) {
+        arr[i] = fn(arr[i], i);
+    }
+    return arr;
+}