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In this post, you will find the solution for the **Add Two Numbers** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Add Two Numbers– LeetCode Problem

Add Two Numbers – LeetCode Problem

**Problem:**

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.

Add Two Numbers– LeetCode Solutions

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; } };

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; } }

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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