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Problem definition is also available here.

Problem summary:

• We have two linked lists as input each stored the digits of a number in reverse order

i.e 107 represented as 7 ➝ 0 ➝ 1

• These input numbers, can be of different length

i.e 107 == 7 ➝ 0 ➝ 1

​ 2099 == 9 ➝ 9 ➝ 0 ➝ 2

• The expected output for the above case is in the same manner

2206 == 6 ➝ 0 ➝ 2 ➝ 2

## Visualize

Let's rewrite these numbers here and flip the arrows.

``````107   ==     1 ← 0 ← 7

2099  == 2 ← 0 ← 9 ← 9
_______________________
2206  == 2 ← 2 ← 0 ← 6

``````

If you see clearly, that's how we compute sum on a paper. Starting from right, carry over when needed and compute the sum. See this special case where result length exceeds the input.

``````                  // carry over
1  ==            1
99  ==      9  ←  9
100  == 1 ←  0  ←  0
``````

Based on above input, we just have to add one decimal position at a time, and pass the carry to the next decimal position. For result, we have to keep chaining next pointer with new digit.

Am approaching this problem with recursive function and a seed — head node. This node will be used to store carry value. Let's iterate on `1 + 99` and see how it looks.

Initial call made [L1, L2, seed0] // Here L1 & L2 are pointers to the input linked lists.

``````initial values

 // seed0
1  ==            1
99  ==      9  ←  9
``````

Below operaitions performed here

1. [seed0.value] + 1 + 9 computed and divided by 10

2. Remainder updated to seed0#value

3. Seed1 created with quotient as the value

4. Seed1 placed as seed0.next

5. Over to next pass [L1.next, L2.next, seed1]

``````pass-1

    // seed1
1  ==            1
99  ==      9  ←  9
0  ==         ←  0
``````

Repeat the same operations here

1. L1 is null now, p2 = 9 and seed1 = 1 so sum is 10. Which is again divided by 10.
2. Remainder is updated to seed1#value
3. Seed2 created takes the carry(1)
4. Seed1#next points to seed2
5. Over to next pass [L1.next, L2.next, seed2]
``````pass-2
           // seed2
1  ==            1
99  ==      9  ←  9
00  ==      0  ←  0
``````
1. L1 = null, p2 = null, seed2 = 1. sum is 1 now.
2. Remainder updated in seed2
3. There is no carry also L1.next & L2.next is null. So, break the recursion here.
``````output:
1  ==            1
99  ==      9  ←  9
________________________
100  == 1 ←  0  ←  0
``````

## Solution

Here is my kotlin solution. I added few best-case checks to avoid looping through the linked list.

``````class ListNode(var `val`: Int) {
var next: ListNode? = null
}

// It is painful to work without an extension
fun ListNode?.valueOrZero() = this?.`val` ?: 0

class Solution {
fun addTwoNumbers(l1: ListNode?, l2: ListNode?): ListNode? {
return when {
l1 == null && l2 == null -> null
l1 == null -> l2
l2 == null -> l1
else -> {
val seed = ListNode(0)
sumInternal(l1, l2, seed)
seed
}
}
}

private fun sumInternal(l1: ListNode?, l2: ListNode?, seed: ListNode) {
val current = seed.`val` + l1.valueOrZero() + l2.valueOrZero()
seed.`val` = current % 10
val carry = current / 10
// Whether input list has any more elements
var hasNext = l1?.next != null || l2?.next != null
if (carry > 0 || hasNext) {
// Create & link the carry here
var newNode = ListNode(carry)
seed.next = newNode
// If there is no digits further -- break here
if (hasNext) {
sumInternal(l1?.next, l2?.next, newNode)
}
}
}
}
``````