# Add Two Numbers Leetcode Solution | C++, Java, Python | Easy Approach

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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 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:

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

Example 2:

```Input: l1 = [0], l2 = [0]
Output: [0]
```

Example 3:

```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.

Time: O(n)
Space: O(1)

### C++

``````class Solution {
public:
ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
ListNode dummy(0);
ListNode* curr = &dummy;
int carry = 0;

while (l1 || l2 || carry) {
if (l1) {
carry += l1->val;
l1 = l1->next;
}
if (l2) {
carry += l2->val;
l2 = l2->next;
}
curr->next = new ListNode(carry % 10);
carry /= 10;
curr = curr->next;
}

return dummy.next;
}
};
``````

### Java

``````class Solution {
public int[] twoSum(int[] nums, int target) {
Map<Integer, Integer> numToIndex = new HashMap<>();

for (int i = 0; i < nums.length; ++i) {
if (numToIndex.containsKey(target - nums[i]))
return new int[] {numToIndex.get(target - nums[i]), i};
numToIndex.put(nums[i], i);
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode(0);
ListNode curr = dummy;
int carry = 0;

while (l1 != null || l2 != null || carry > 0) {
if (l1 != null) {
carry += l1.val;
l1 = l1.next;
}
if (l2 != null) {
carry += l2.val;
l2 = l2.next;
}
curr.next = new ListNode(carry % 10);
carry /= 10;
curr = curr.next;
}

return dummy.next;
}
}
``````

### Python

``````class Solution:
class Solution:
def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
dummy = ListNode(0)
curr = dummy
carry = 0

while carry or l1 or l2:
if l1:
carry += l1.val
l1 = l1.next
if l2:
carry += l2.val
l2 = l2.next
curr.next = ListNode(carry % 10)
carry //= 10
curr = curr.next

return dummy.next

``````

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