LeetCode 2. Add Two Numbers 链表加和问题

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)

Output: 7 -> 0 -> 8

题目解释：两个链表按顺序依次加和，有进位的向后进位，得到结果

public ListNode addTwoNumbers(ListNode l1, ListNode l2) {int count = 0;    ListNode result = new ListNode(0);    ListNode p = result;while(l1 != null||l2 != null){if( l1 != null){count += l1.val;l1 = l1.next;}if( l2 != null){count += l2.val;l2 = l2.next;}ListNode kk = new ListNode(count%10);p.next = kk;p = p.next;count /= 10;}if(count!=0){ListNode n1 = new ListNode(count);p.next = n1;}    return result.next;}

注意的点：

1.定义返回结果result时，需要将result锁住，再指定一个p指向result，p依次向后迭代，最重要的是p = p.next 一定要移动

2.初始写代码时定义了进位符flag，但是会出现没有进位会自动创建一个为0的ListNode会与题意不符，直接创建count%10的ListNode，返回result.next

错误代码：

flag = (result.val + count)/10;ListNode next = new ListNode(flag);