题目链接：https://leetcode.com/problems/add-two-numbers/description/
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
思路：
遍历两个链表，每位数字累加（同时需要加上结果链表该节点的值），模10为结果为结果列表当前指针值，模10为结果链表next指针初始值。
注意事项：
1.结束条件，除10只要大于0，next指针就需要新建。而不是l1，l2遍历完了，就不新建next指针了
if(key > 0){
curr.next = new ListNode(key);
}
不是
if(p != null && q != null){
curr.next = new ListNode(key);
}
2.两个链表长度不相等的情况,遍历l1,l2的结束条件是 “或”
examples:
l1=[1,2];
l2=[3];
java代码
public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode p = l1;
ListNode q = l2;
ListNode l3 = new ListNode(0);
ListNode curr = l3;
int key = 0;
while (p != null || q != null) {
int pV = (p != null) ? p.val : 0;
int qV = (q != null) ? q.val : 0;
int tmp = pV + qV + key;
key = tmp / 10;
int mo = tmp % 10;
curr.next = new ListNode(mo);
curr = curr.next;
p = (p != null) ? p.next : p;
q = (q != null) ? q.next : q;
if(key > 0){
curr.next = new ListNode(key);
}
}
return l3.next;
}
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
