Add solution for 2. Add Two Numbers

Author: vkazakevich

README.md

@@ -9,6 +9,7 @@ The repository contains the best versions of my solutions to [LeetCode](https://
 | # | Title | Solution | Time | Space |
 |---| ----- | -------- | ---- | ----- |
 |1814|[Count Nice Pairs in an Array](https://leetcode.com/problems/contains-duplicate/)|[TypeScript](./Solutions/TypeScript/1814-count-nice-pairs-in-an-array.ts)|O(n)|O(n)|
+|2|[Add Two Numbers](https://leetcode.com/problems/add-two-numbers/)|[TypeScript](./Solutions/TypeScript/2-add-two-numbers.ts)|O(n)|O(n)|
 
 ## Solutions Python
 

Solutions/TypeScript/2-add-two-numbers.ts

@@ -0,0 +1,65 @@
+/** 
+You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+Example 1:
+Input: l1 = [2,4,3], l2 = [5,6,4]
+Output: [7,0,8]
+Explanation: 342 + 465 = 807.
+
+Example 2:
+Input: l1 = [0], l2 = [0]
+Output: [0]
+Example 3:
+
+Input: l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]
+Output: [8,9,9,9,0,0,0,1]
+ 
+Constraints:
+
+The number of nodes in each linked list is in the range [1, 100].
+0 <= Node.val <= 9
+It is guaranteed that the list represents a number that does not have leading zeros.
+*/
+
+class ListNode {
+  val: number;
+  next: ListNode | null;
+  constructor(val?: number, next?: ListNode | null) {
+    this.val = val === undefined ? 0 : val;
+    this.next = next === undefined ? null : next;
+  }
+}
+
+function addTwoNumbers(
+  l1: ListNode | null,
+  l2: ListNode | null
+): ListNode | null {
+  let current = new ListNode(0);
+  let result = current;
+
+  let carry = 0;
+
+  while (l1 || l2 || carry) {
+    let value1 = l1 ? l1.val : 0;
+    let value2 = l2 ? l2.val : 0;
+
+    let sum = value1 + value2 + carry;
+
+    carry = 0;
+
+    if (sum > 9) {
+      carry = ~~(sum / 10);
+      sum = sum % 10;
+    }
+
+    current.next = new ListNode(sum);
+    current = current.next;
+
+    if (l1) l1 = l1.next;
+    if (l2) l2 = l2.next;
+  }
+
+  return result.next;
+}