LeetCode 2. Add Two Numbers [Algorithm + Code Explained ] Best Solution

One of the most frequently asked coding interview questions on Arrays in companies like Google, Facebook, Amazon, LinkedIn, Microsoft, Uber, Apple, Adobe etc.

LeetCode : Add Two Numbers

Question - 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 contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 - 4 - 3) + (5 - 6 - 4)
Output: 7 - 0 - 8
Explanation: 342 + 465 = 807

[hiteshmoorjani]

Great explanation, A bit optimization in above code, We are adding only single-digit of LinkedList, thus we can tell the carry would not be greater than 1(Maximum 1 digit calculation would be (9+9) which 18 and carry is 1).

if(sum >= 10)
{
carry = 1;
sum = sum%10;
}



And also we can put the condition only perform carry operation when the sum is greater than or equal to 10.

[AbhinavKumar-dref]

Awesome Jayati your concept is so clear how the new node is created and pointer is moved to the next node. Thank you for a good and simple approach. Keep this good work.

[ryanbennett]

You explain these concepts really well!

[mkiruba]

Great videos, It would be nice if there is a link to the slide shown in 2:20

[mohammedsaleh]

5722765 - 3000 what type of algorithm is this? Please help!

[indranilthakur]

Why are we returning sumHead.next and not sumHead?

[gauravshukla]

perfect example of beauty with brain...😅..thanks for the concept

[imagination]

Jitne log amswer nhi deak rahe honge ushe zayda to jo padha rahi hai use deak rahe hone

😅😅😅

[laptewstein]

implementation in ruby
```
def add_two_numbers(l1, l2)
pointer_to_head = ListNode.new # dummy
tail = pointer_to_head
local_sum = 0
until l1.nil? && l2.nil? && local_sum == 0
if l1
local_sum += l1.val
l1 = l1.next
end
if l2
local_sum += l2.val
l2 = l2.next
end
tail.next = ListNode.new local_sum % 10 # hook up
local_sum = local_sum / 10 # serves as a carryover
tail = tail.next # reassign running pointer to the newly created node
end
pointer_to_head.next
end

```