题目描述：
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 contains a single digit. Add the two numbers and return the sum as a linked list.
You may assume the two numbers do not contain any leading zero, except the number 0 itself.
Example 1:
Input: l1 = [2,4,3], l2 = [5,6,4]
Output: [7,0,8]
Explanation: 342 + 465 = 807.
Example 2:
Input: l1 = [0], l2 = [0]
Output: [0]
Example 3:
Input: nums = [3,3], target = 6
Output: [0,1]
Constraints:
The number of nodes in each linked list is in the range [1, 100].
0 <= Node.val <= 9
It is guaranteed that the list represents a number that does not have leading zeros.
Time complexity: O(n+m)
模拟加法运算：
1、同时从头开始枚举两个链表，sum 的值等于l1和l2链表的当前元素的和加上之前进位 carry的值，再将 sum% 10 的元素存到dummy链表中，将sum / 10的值赋给carry，l1和l2再同时往后移动
2、循环终止条件。当遍历完所有元素并且 carry == 0。
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode() {}
* ListNode(int val) { this.val = val; }
* ListNode(int val, ListNode next) { this.val = val; this.next = next; }
* }
*/
class Solution {
public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
ListNode dummy = new ListNode();
ListNode p = l1, q = l2, curr = dummy;
int carry = 0;
while(l1 != null || l2 != null || carry != 0){
int x = l1 == null ? 0:l1.val;
int y = l2 == null ? 0:l2.val;
int sum = x + y + carry;
carry = sum /10;
curr.next = new ListNode(sum % 10);
if(l1 != null) l1 = l1.next;
if(l2 != null) l2 = l2.next;
curr = curr.next;
}
return dummy.next;
}
}