LeetCode Problem: Linkded List Topics

1. 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. For example:

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

Suppose the input lists are:

l1: 2 -> 4 -> 3 -> 2
l2: 5 -> 6 -> 4 -> null

The initial carry is set to 0, and we can use two pointers to traverse the two lists at the same time:

2 -> 4 -> 3 -> 2
⬆

5 -> 6 -> 4 -> null
⬆

2 + 5 + carry = 7, and the updated carry is 0. We can add 7%10 = 7 into the result list, and keep traversing the two lists:

2 -> 4 -> 3 -> 2
     ⬆

5 -> 6 -> 4 -> null
     ⬆

4 + 6 + carry = 10, and the updated carry should be 1. We can add 10%10 = 0 into the result list, and keep traversing the two lists:

2 -> 4 -> 3 -> 2
          ⬆

5 -> 6 -> 4 -> null
          ⬆

3 + 4 + carry = 8, and the updated carry is 0. We can add 8%10 = 8 into the result list, and keep traversing the two lists:

2 -> 4 -> 3 -> 2
               ⬆

5 -> 6 -> 4 -> null
               ⬆

2 + 0 + carry = 2, and the updated carry is 0. We can add 2%10 = 2 into the result list, and keep traversing the two lists.

After traversing the two lists, the final carry is 0, therefore final result list is 7 -> 0 -> 8 -> 2.

Here is the code:

public class AddTwoNumbers {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode head = new ListNode(-1);// the head node of the result list
        ListNode current = head;

        ListNode p1 = l1;// pointer1
        ListNode p2 = l2;// pointer2
        int carry = 0;

        while (p1 != null || p2 != null) {

            int sum = (p1 == null ? 0 : p1.val)
                    + (p2 == null ? 0 : p2.val)
                    + carry;

            // update carry
            if (sum >= 10) {
                sum %= 10;
                carry = 1;
            } else {
                carry = 0;
            }

            current.next = new ListNode(sum);
            current = current.next;

            // update the pointers
            p1 = (p1 == null ? null : p1.next);
            p2 = (p2 == null ? null : p2.next);
        }

        // do not forget to check the final carry
        // for example, if we have:
        // l1: 2 -> 4 -> 3 -> 2
        // l2: 5 -> 6 -> 4 -> 9
        // the final carry is 1 and we need to add 1 into the 
        // result list
        if (carry == 1) {
            current.next = new ListNode(1);
        }

        return head.next;
    }
}

class ListNode {
    int val;
    ListNode next;

    ListNode(int x) {
        val = x;
    }
}