LeetCode: 2_Add Two Numbers | 两个链表中的元素相加 | Medium

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

 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 // }