Description
You have two numbers represented by a linked list, where each node contains a single digit. The digits are stored in reverse order, such that the 1's digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.
Example
Given 7->1->6 + 5->9->2. That is, 617 + 295.
Return 2->1->9. That is 912.
Given 3->1->5 and 5->9->2, return 8->0->8
解题：题目的意思是，给两个链表，倒序表示一个若干位的数。要求返回这两个数的和，并且格式是题中规定的链表，也是倒序。思路很清晰，从头到尾，一位一位地相加，并且用一个数来保存进位。每次相加的时候，都得考虑进位，把进位算在其中。当然，思路越清晰，代码可能看起来就比较笨重。代码如下：
1 /**
2 * Definition for ListNode
3 * public class ListNode {
4 * int val;
5 * ListNode next;
6 * ListNode(int x) {
7 * val = x;
8 * next = null;
9 * }
10 * }
11 */
12
13 public class Solution {
14 /**
15 * @param l1: the first list
16 * @param l2: the second list
17 * @return: the sum list of l1 and l2
18 */
19 public ListNode addLists(ListNode l1, ListNode l2) {
20 // write your code here
21 ListNode head = new ListNode(0);
22 ListNode p=head;
23 ListNode p1 = l1;
24 ListNode p2 = l2;
25 int sum = 0;//保存每一位的和
26 int more=0;//保存进位
27 while(p1 != null && p2 != null){
28 sum = p1.val + p2.val+more;
29 more = sum / 10;
30 sum = sum % 10;
31 ListNode temp = new ListNode(sum);
32 p.next = temp;
33 p=p.next;
34 p1 = p1.next;
35 p2 = p2.next;
36 }
37 //如果more不为0
38 sum = 0;
39 while(p1 != null){
40 sum = more + p1.val;
41 more = sum / 10;
42 sum = sum % 10;
43 ListNode temp = new ListNode(sum);
44 p.next = temp;
45 p = p.next;
46 p1 = p1.next;
47 }
48 while(p2 != null){
49 sum = more + p2.val;
50 more = sum / 10;
51 sum = sum % 10;
52 ListNode temp = new ListNode(sum);
53 p.next = temp;
54 p = p.next;
55 p2 = p2.next;
56 }
57 if(more != 0){ //如果more不为0，那么最高位就是前面的进位
58 ListNode temp = new ListNode(more);
59 p.next = temp;
60 }
61 return head.next;
62 }
63 }
代码可能有点冗余，曾尝试改进，但发现还是这样写看起来比较清晰。如有错误，欢迎批评指正。