### 描述​

You are given two non-empty linked lists representing two non-negative integers. The most significant digit comes first 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.

Example:

Input: (7 -> 2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 8 -> 0 -> 7

// Add Two Numbers II// Time Complexity: O(m+n), Time Complexity: O(1)public class Solution {    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {        l1 = reverseList(l1);        l2 = reverseList(l2);        ListNode result = addTwoNumbersI(l1, l2);        return reverseList(result);    }    private static ListNode addTwoNumbersI(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;    }    private static ListNode reverseList(ListNode head) {        ListNode p = null;        ListNode q = head;        while (q != null) {            ListNode tmp = q.next;            q.next = p;            p = q;            q = tmp;        }        return p;    }};