## Description

LeetCode Problem 2.

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:

``````1
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Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
``````

Example 2:

``````1
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Input: l1 = , l2 = 
Output: 
``````

Example 3:

``````1
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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.

## Sample C++ Code

This is a C++ solution using a `hash table`.

``````1
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/**
* struct ListNode {
*     int val;
*     ListNode *next;
*     ListNode() : val(0), next(nullptr) {}
*     ListNode(int x) : val(x), next(nullptr) {}
*     ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
//result
ListNode *res = new ListNode();
int carryOver = 0;
int val1 = 0;
int val2 = 0;

while((l1 != NULL) || (l2 != NULL)) {
if((l1 != NULL) && (l2 != NULL)) {
//we have both values
val1 = l1->val;
val2 = l2->val;
//increment both
l1 = l1->next;
l2 = l2->next;
}
else if((l1 != NULL) && (l2 == NULL)) {
//we have one value
val1 = l1->val;
val2 = 0;
//increment l1
l1 = l1->next;
}
else if((l1 == NULL) && (l2 != NULL)) {
//we have one value
val1 = 0;
val2 = l2->val;
//increment l2
l2 = l2->next;
}

int sum = val1 + val2 + carryOver;
res->val = sum % 10;
carryOver = sum / 10;

if(l1 != NULL || l2 != NULL || carryOver > 0) {
ListNode* nextNode = new ListNode();
res->next = nextNode;
res = res->next;
}
}

if(carryOver > 0)
res->val = carryOver;

}
};
``````