LeetCode -- Add Two Numbers

LeetCode -- Add Two Numbers
Description:


You are given two linked lists representing two non-negative numbers. 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.


Input: (2-> 4-> 3) + (5-> 6-> 4)
Output: 7-> 0-> 8


At the same time, the two linked lists are traversed, and each node is summed separately. Each node stores only one result, and the carry is reserved for the calculation of the next node.


Ideas:
1. The two pointers point to the first place of the linked list 1 (l1) and linked list 2 (l2) respectively, and can be calculated by bit.
2. Save the first node and when the linked list is traversed, there is still a situation where carry is not placed in the linked list.

/** * Definition for singly-linked list. * public class ListNode { *     public int val; *     public ListNode next; *     public ListNode(int x) { val = x; } * } */public class Solution {    public ListNode AddTwoNumbers(ListNode l1, ListNode l2) {        ListNode node = null;            ListNode head = null;        var carry = 0;    while(l1 != null || l2 != null){        var a = l1 != null ? l1.val : 0;    var b = l2 != null ? l2.val : 0;        var s = a + b + carry;    var r = s % 10;    if(node == null){    node = new ListNode(r);    head = node;    }else{    node.next = new ListNode(r);    node = node.next;    }    carry = s / 10;        if(l1 != null){    l1 = l1.next;    }    if(l2 != null){    l2 = l2.next;    }    }        if(carry > 0){    var n = new ListNode(carry);    node.next = n;    }        return head;    }}