算法#25--两整数链表求和

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

My Code:

/**
* 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 first = null;
ListNode last = null;
int v = 0;
while (true)
{
ListNode oldlast = last;
if (l1 == null && l2 == null && v == 0)
{
break;
}

if (l1 == null)
{
l1 = new ListNode(0);
}

if (l2 == null)
{
l2 = new ListNode(0);
}

int s = l1.val + l2.val + v;
v = s / 10;
last = new ListNode(s % 10);
l1 = l1.next;
l2 = l2.next;

if (first == null)
{
first = last;
}
else
{
oldlast.next = last;
}
}
return first;
}
}

Best Code:

public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode c1 = l1;
ListNode c2 = l2;
ListNode sentinel = new ListNode(0);
ListNode d = sentinel;
int sum = 0;
while (c1 != null || c2 != null) {
sum /= 10;
if (c1 != null) {
sum += c1.val;
c1 = c1.next;
}
if (c2 != null) {
sum += c2.val;
c2 = c2.next;
}
d.next = new ListNode(sum % 10);
d = d.next;
}
if (sum / 10 == 1)
d.next = new ListNode(1);
return sentinel.next;
}
}