### 描述​

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

### 分析​

// Add Two Numbers// 跟Add Binary 很类似// 时间复杂度O(m+n)，空间复杂度O(1)public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode dummy = new ListNode(-1); // 头节点        int carry = 0;        ListNode prev = dummy;        for (ListNode pa = l1, pb = l2;             pa != null || pb != null;             pa = pa == null ? null : pa.next,             pb = pb == null ? null : pb.next,             prev = prev.next) {            final int ai = pa == null ? 0 : pa.val;            final int bi = pb == null ? 0 : pb.val;            final int value = (ai + bi + carry) % 10;            carry = (ai + bi + carry) / 10;            prev.next = new ListNode(value); // 尾插法        }        if (carry > 0)            prev.next = new ListNode(carry);        return dummy.next;    }};