Add Two Numbers in Linked List Leet Code 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.

/**

 * Definition for singly-linked list.

 * public class ListNode {

 *     int val;

 *     ListNode next;

 *     ListNode() {}

 *     ListNode(int val) { this.val = val; }

 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }

 * }

 */

class Solution {

    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {

       

         ListNode dummyHead = new ListNode(0); // Step 1: Create a dummy node

        ListNode current = dummyHead;  // Pointer to build the new list

        int carry = 0;  // Step 2: Initialize carry

        while (l1 != null || l2 != null || carry > 0) {  // Step 3: Traverse both lists

            int x = (l1 != null) ? l1.val : 0;  // Get value of l1 (or 0 if null)

            int y = (l2 != null) ? l2.val : 0;  // Get value of l2 (or 0 if null)

            int sum = x + y + carry;  // Step 4: Compute sum

            carry = sum / 10;  // Step 5: Calculate carry

            current.next = new ListNode(sum % 10); // Step 6: Create a new node for result

            current = current.next;  // Move the pointer forward

            if (l1 != null) l1 = l1.next;  // Move l1 forward

            if (l2 != null) l2 = l2.next;  // Move l2 forward

        }

        return dummyHead.next;  // Step 7: Return result (skip dummy node)

    }

}