跟我一起学算法系列2---Add Two Numbers（jav

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;
    }
}