题目：
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
解题思路：
两个链表中的元素相加相比普通的加法操作增加了一个层次，和两个数组中的元素相加意义上差不多，这个题可以引申到大数相加上。需要注意的是：进位的处理。
代码展示：
1 #include <iostream>
2 #include <cassert>
3
4 using namespace std;
5
6
7 //Definition for singly-linked list.
8 struct ListNode {
9 int val;
10 ListNode *next;
11 ListNode(int x) : val(x), next(NULL) {}
12 };
13
14 ListNode * insertNodes(int n)
15 {
16 ListNode *pNode = new ListNode(0);
17 ListNode *p = pNode;
18 int i = 0;
19 int m;
20 while(i < n) {
21 cin >> m;
22 ListNode *pIn = new ListNode(m);
23 p->next = pIn;
24 p = pIn;
25
26 i++;
27 }
28 return pNode->next;
29 }
30
31 void Print(ListNode *p)
32 {
33 ListNode *pT = p;
34 while(NULL != pT) {
35 cout << pT->val << " ";
36 pT = pT->next;
37 }
38 }
39
40 ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
41 {
42 assert(NULL != l1);
43 assert(NULL != l2);
44
45 ListNode *pRetNode = new ListNode(0);
46 ListNode *p = pRetNode;
47
48 int div = 0;
49 int nSumVal;
50 while (NULL != l1 || NULL != l2) {
51 int val1 = 0, val2 = 0;
52 if (NULL != l1) {
53 val1 = l1->val;
54 l1 = l1->next;
55 }
56 if (NULL != l2) {
57 val2 = l2->val;
58 l2 = l2->next;
59 }
60 nSumVal = val1 + val2 + div; //一个比较巧妙的进位相加的方式
61 div = nSumVal/10;
62 ListNode *pT = new ListNode(nSumVal%10);
63 p->next = pT;
64 p = pT;
65 }
66 if (div) {
67 p->next = new ListNode(div);
68 }
69 return pRetNode->next;
70 }
71
72 // int main()
73 // {
74 // ListNode *p1 = insertNodes(1);
75 // ListNode *p2 = insertNodes(4);
76 // ListNode *p = addTwoNumbers(p1,p2);
77 // Print(p);
78 // return 0;
79 // }