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.
Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
题意：
给你两个表示两个非负数字的链表。数字以相反的顺序存储，其节点包含单个数字，将这两个数字相加并将其作为一个链表返回。
解法：
需要注意的点：
1）两个list可能不一样长
2）两个digits 相加如果大于等于10，需要进位
3）输出的node要reverse（反向）
实现：
public class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
int carry = 0;
ListNode newNode = new ListNode(0);
ListNode p1 = l1;
ListNode p2 = l2;
ListNode p3 = newNode;
while(null != p1 || null != p2)
{
if(null != p1)
{
carry += p1.val;
p1 = p1.next;
}
if(null != p2)
{
carry += p2.val;
p2 = p2.next;
}
p3.next = new ListNode(carry%10);
p3 = p3.next;
carry /= 10;
}
if(carry == 1)
p3.next = new ListNode(1);
return newNode.next;
}
}