## Question

https://leetcode.com/problems/add-two-numbers/

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.

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

## Explanation

This problem test the skills of linked list node operation. There are two points here:

- Merge two linkedl ist into one
- Use the dummy node to simplify code logic

### Merge two linked lists

It’s straightforward to iterate over the two link node pointers. But please notice the scenario of one linked list is longer than the other one, so we need to test whenever we need to fetch data from a list node. This is an error-prone point for starter.

Assume that \(m\) and \(n\) represents the length of \(l1\) and \(l2\) respectively:

The time complexity: \(O(max(M, N))\)

The space complexity: \(O(max(M, N))\)

### Use the dummy node

What’s `dummy node`

?

If we don’t use `dummy node`

, we must have a variable which pointing to the final result. If we don’t initialize it beforehand, we need to construct it during the loop. the pseudo code should be:

node* res = null node* p = null while(...) { if(res == null) { res = new node with val p = res } .... }

Hence the code is a little complicated, and it’s also mixed with other logic for node merging.

The idea of `dummy node`

is we initialize a unused node firstly, then we always keep the dummy.next to be the head node.

This is good taste for linked list.

## C++

```
class Solution {
public:
ListNode *addTwoNumbers(ListNode *l1, ListNode *l2) {
ListNode dummy(0);
ListNode* p = &dummy;
int carry = 0, sum;
while(l1 || l2) {
sum = carry;
if(l1) sum += l1->val;
if(l2) sum += l2->val;
if(sum >= 10) {
carry = sum / 10;
sum %= 10;
} else carry = 0;
p->next = new ListNode(sum);
p = p->next;
if(l1) l1 = l1->next;
if(l2) l2 = l2->next;
}
if(carry)
p->next = new ListNode(carry);
return dummy.next;
}
};
```

## Java

Notice this Java version don’t test the sum >= 10, I keep it in C++ version for a little better time performance.

```
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummyHead = new ListNode(0);
ListNode p = l1, q = l2, curr = dummyHead;
int carry = 0;
while (p != null || q != null) {
int x = (p != null) ? p.val : 0;
int y = (q != null) ? q.val : 0;
int sum = carry + x + y;
carry = sum / 10;
curr.next = new ListNode(sum % 10);
curr = curr.next;
if (p != null) p = p.next;
if (q != null) q = q.next;
}
if (carry > 0) {
curr.next = new ListNode(carry);
}
return dummyHead.next;
}
```

## Python

```
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
resHead = resList = ListNode(None)
carry = 0
while l1 or l2 :
sum_val = carry
if l1 != None:
sum_val += l1.val
l1 = l1.next
if l2 != None:
sum_val += l2.val
l2 = l2.next
carry = sum_val // 10
resList.next = ListNode(sum_val % 10)
resList = resList.next
if carry == 1:
resList.next = ListNode(carry)
return resHead.next
```