【leetcode】Add Two Numbers

• 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.
  Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

java解法：
链表结构，倒序求和，只要在求完和后判断是否大于10，如果大于10，置carry标签为1，并在新的链表中加上carry，注意：三位数+三位数=可能为四位数，所以需要在最后判断carry存在否new新的点。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(-1);
        ListNode cur = dummy;
        int carry = 0;
        while (l1 != null || l2 != null) {
            int d1 = l1 == null ? 0 : l1.val;
            int d2 = l2 == null ? 0 : l2.val;
            int sum = d1 + d2 + carry;
            carry = sum >= 10 ? 1 : 0;
            cur.next = new ListNode(sum % 10);
            cur = cur.next;
            if (l1 != null) l1 = l1.next;
            if (l2 != null) l2 = l2.next;
        }
        if (carry == 1) cur.next = new ListNode(1);
        return dummy.next;
    }
}