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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0001_0100.s0002_add_two_numbers;

// #Medium #Top_100_Liked_Questions #Top_Interview_Questions #Math #Linked_List #Recursion
// #Data_Structure_II_Day_10_Linked_List #Programming_Skills_II_Day_15
// #Big_O_Time_O(max(N,M))_Space_O(max(N,M)) #2023_08_09_Time_1_ms_(100.00%)_Space_43.1_MB_(78.48%)

import com_github_leetcode.ListNode;

/*
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
/**
 * 2 - Add Two Numbers\.
 *
 * Medium
 *
 * 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:**
 *
 * ![](https://assets.leetcode.com/uploads/2020/10/02/addtwonumber1.jpg)
 *
 * **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.
**/
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummyHead = new ListNode(0);
        ListNode p = l1;
        ListNode q = l2;
        ListNode curr = dummyHead;
        int carry = 0;
        while (p != null || q != null) {
            int x = (p != null) ? p.val : 0;
            int y = (q != null) ? q.val : 0;
            int sum = carry + x + y;
            carry = sum / 10;
            curr.next = new ListNode(sum % 10);
            curr = curr.next;
            if (p != null) {
                p = p.next;
            }
            if (q != null) {
                q = q.next;
            }
        }
        if (carry > 0) {
            curr.next = new ListNode(carry);
        }
        return dummyHead.next;
    }
}