LeetCode 2.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.

思路：链表合并与实现加法运算的结合版。有三种情况：
1.两个链表都不为空同时判断是否有进位，设置标志位，取余，处除10
2.一个链表为空另一个不为空同时判断是否有进位，若有进位与1判断相同，若无进位直接将链接追加到新链表尾部。
3.两个链表都为空同时判断是否有进位，有进位在尾部追加1。

 1 public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
 2         ListNode head = null;
 3         ListNode rear = null;
 4         int flag = 0;
 5         ListNode node = null;
 6         while(l1!=null&&l2!=null) {//处理两个链表对应位置都不为0的情况
 7             if(flag==1) {//是否进位
 8                 node = new ListNode((l1.val+l2.val+1)%10);
 9             }else {
10                 node = new ListNode((l1.val+l2.val)%10);
11             }//进位
12             flag = (l1.val+l2.val+flag)/10;
13             if(head==null) {//头节点为空
14                 head = node;
15                 rear = head;
16             }else {
17                 rear.next = node;
18                 rear = rear.next;
19             }
20             l1 = l1.next;
21             l2 = l2.next;
22         }
23         //有进位需要进行特殊处理。比如5+5=10，1+99=100。
24         while(flag==1) {//有一个链表对应位置为空,同时有进位
25             if(l1==null&&l2==null) {
26                 node = new ListNode(1);
27                 flag = 0;
28             }
29             else {
30                 if(l1!=null) {
31                     node = new ListNode((l1.val+1)%10);
32                     flag = (l1.val+1)/10;
33                     l1 = l1.next;
34                 }
35                 if(l2!=null) {
36                     node = new ListNode((l2.val+1)%10);
37                     flag = (l2.val+1)/10;
38                     l2 = l2.next;
39                 }
40             }
41             rear.next = node;
42             rear = rear.next;
43         }
44         //无进位，只需要把l1或l2剩下的节点追加到新链表中
45         while(l1!=null) {
46             rear.next = l1;
47             rear = rear.next;
48             l1 = l1.next;
49         }
50         while(l2!=null) {
51             rear.next = l2;
52             rear = rear.next;
53             l2 = l2.next;
54         }
55         return head;
56         
57     }