LeetCode 2. Add Two Numbers

原题题目： 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

一开始题目都没看懂，题目的意思是给你两个链表，里边装有倒序的数字，例如2->4->3对应的是342，5->6->4对应的是465，输出是这两数字相加的结果倒序：7->0->9（ 342+465=907）

题目本身并不难，只要按照我我们平时做加法的顺序就行了。超过十向高位进1，高位没数字则置1。

从该题目可以学习的一些技巧：

1.如何使用头结点，返回头结点。

2.如何用一个while循环遍历完l1和l2

代码：

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode l3 = new ListNode(0);        ListNode head = l3;        int sum = 0;        while (l1 !=null || l2 != null) {            sum = sum>9 ? 1 : 0;            if (l1 != null) {                sum += l1.val;                l1 = l1.next;            }            if (l2 != null) {                sum += l2.val;                l2 = l2.next;            }            l3.next = new ListNode(sum%10);            l3 = l3.next;        }        if (sum > 9) {            l3.next = new ListNode(1);        }        return head.next;    }}