2. 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.

给定两个非空链表，表示两个非负整数。数字是倒序存储的，每个节点包括一个单独的数字。将这两个数字相加，把结果以链表的形式返回。

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

可以假设这两个数字均不以0开头，除了0本身。

Example

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

这个问题和67. Add Binary比较像~

potpie的做法

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;
    }
}

我自己的做法，大致和高票答案一样，只是我的把结果链表头单独计算，代码复杂了一点

if(l1==null)
return l2;
if(l2==null)
return l1;

int carry=0;
int sum=0;

ListNode result=new ListNode((l1.val+l2.val)%10);
carry=(l1.val+l2.val)/10;;
l1=l1.next;
l2=l2.next;
ListNode pointer=result;

while(l1!=null||l2!=null)
{
sum=carry;
if(l1!=null)
{
sum=sum+l1.val;
l1=l1.next;
}
if(l2!=null)
{
sum=sum+l2.val;
l2=l2.next;
}

carry=sum/10;
pointer.next=new ListNode(sum%10);
pointer=pointer.next;

}
if (carry!=0)
pointer.next=new ListNode(carry);
return result;