Leetcode第二题_Add Two Numbers

Add Two Numbers Total Accepted: 58281 Total Submissions: 266931 My Submissions Question Solution
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

这一题应该算是比较简单的题目，题目大概的意思是给你两个链表，将这两个链表中对应位置的节点求和，如果该处和大于10则进位，然后输出和链表。

思路使用尾插法新建出一个和链表就可以了。思路是这么简单，但这一题却有许多特殊情况要考虑

1. 一个链表长一个链表短的情况
2. 最后一个节点处有进位的情况
3. 不等长链表最后一个节点处有进位的情况

处理好这三种情况就差不多AC了。下面是代码：

 public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {        ListNode temp = new ListNode(0);//      记录下头结点，尾插法构建和链表        ListNode head = temp;//      进位标记        int flag = 0;        int delete = 0;        while (l1!= null || l2!= null) {//          处理链表不等长            if (l1 == null) {//              处理进位                l2.val = l2.val+flag;                flag = 0;//              处理不等长节点最后进位情况                if (l2.val >=10) {                    delete = -10;                    flag = 1;                }                ListNode l3 = new ListNode(l2.val+delete);                delete = 0;                temp.next = l3;                temp = l3;                l2 = l2.next;            }            else if (l2 == null) {                l1.val = l1.val + flag;                flag = 0;                if (l1.val >=10) {                    delete = -10;                    flag = 1;                }                ListNode l3 = new ListNode(l1.val+delete);                delete = 0;                temp.next = l3;                temp = l3;                l1 = l1.next;            }            else {                l1.val = l1.val + flag;                flag = 0;                if (l1.val+l2.val>=10) {                    delete = -10;                    flag = 1;                }                ListNode l3 = new ListNode(l1.val+l2.val+delete);                delete = 0;                temp.next = l3;                temp = l3;                l1 = l1.next;                l2 = l2.next;            }//          处理最后一个节点有进位情况            if (flag ==1 && l1 ==null && l2 ==null) {                ListNode l3 = new ListNode(1);                temp.next = l3;                temp = l3;            }        }        temp.next = null;        return head.next;    }

其实挺烦的，每刷一个题要2,3个小时，感觉好消耗时机。处理问题的时候总是考虑不全面，这些特例都是通过Leetcode给失败反馈找出来的，要是不给反馈，这些例子怎么找到的？肯定要花更长的时间。

真是烦，实习又没有找到，感觉自己欠缺的还挺多的，哎，时间都去哪了啊。