Add Two Numbers

In this article we will learn about LeetCode Second question Add Two Numbers in detail with optimized solution with Example.

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.

Table of Contents

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.

Java Example

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
    ListNode ans = new ListNode(0);
    ListNode curr = ans;
    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 ans.next;
        
  }
}
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C++ Example

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

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

    return ans.next;
  }
};

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Complexity :

Time complexity: O(N);
Space Complexity: O(1);

I hope you like this Amazing Add Two Numbers article.🙂