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.
Examples :

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

This problem was taken from Leetcode

## Solution

The solution:

### Java Script

```/**
* function ListNode(val, next) {
*     this.val = (val===undefined ? 0 : val)
*     this.next = (next===undefined ? null : next)
* }
*/

function ListNode(val, next) {
this.val = (val===undefined ? 0 : val)
this.next = (next===undefined ? null : next)
}

/**
* @param {ListNode} l1
* @param {ListNode} l2
* @return {ListNode}
*/
var addTwoNumbers = function(l1, l2) {

function doSum(a,b, carry) {
var result = {
val: 0,
carry: 0
}
var sum = a + b + carry;
if(sum > 9) {
result.val = sum - 10;
result.carry = 1;
}
else {
result.val = sum;
result.carry = 0;
}
return result;
}

var nodeA = l1;
var nodeB = l2;
var end = false;
var carry = 0;
var result = new ListNode(0, null);
var resultPointer = result;
var tmp = null;

while(nodeA != null || nodeB != null) {
var valA = nodeA == null ? 0 : nodeA.val;
var valB = nodeB == null ? 0 : nodeB.val;
var sumResult = doSum(valA, valB, carry);
carry = sumResult.carry;

resultPointer.val = sumResult.val;
resultPointer.next = new ListNode(0, null);
tmp = resultPointer;
resultPointer = resultPointer.next;

nodeA = nodeA ? nodeA.next : null;
nodeB = nodeB ? nodeB.next : null;
}

if(carry != 0) {
tmp.next.val = carry; // add carry to the val of the last node
}
else
tmp.next = null; // remove the last empty node.
return result;
};

var l1 = new ListNode(9,
new ListNode(9,
new ListNode(9,
new ListNode(9, null)
))
);

var l2 = new ListNode(9,
new ListNode(9,
new ListNode(9,
new ListNode(9,
new ListNode(9,
new ListNode(9,
new ListNode(9, null)
)))))
)

// [9,9,9] , [9,9,9,9,9]
// result: [8,9,9,0,0,1]