Add two numbers leetcode

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

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Linked List Math

思路分析：

　　（1）首先对两个链表从前往后开始遍历，逐位相加（计为res），则遍历中的当前位为：res对10取摸,进位（计为carry）为：res/10;这样的过程持续到当中任意一个链表为空时结束，此时有3种情况:

第一种：l1为空，l2不空；第二种：l2为空，l1不空；第三种：l1，l2都为空；

　　 (2)针对3种情况分别加以处理，此时一定要注意链表移到末尾但进位为1的情况；

　　（3）边界问题：l1或者l2为空的情况！

代码如下，这次是使用c语言实现：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     struct ListNode *next;
 6  * };
 7  */
 8 struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
 9     if (l1 == NULL)
10         return l2;
11     if (l2 == NULL)
12         return l1;
13     struct ListNode*p1 = l1, *p2 = l2, *pNew = NULL,*pre=p1;
14     int res, carry = 0;
15     while (p1&&p2)
16     {
17         res = p1->val + p2->val + carry;
18         carry = res / 10;
19         p1->val = res % 10;
20         pre = p1;
21         p1 = p1->next;
22         p2 = p2->next;
23     }
24     if (p2 != NULL)
25     {
26         pre->next = p2;
27         while (p2)
28         {
29             res = (p2->val + carry);
30             p2->val = res % 10;
31             carry = res / 10;
32             pre=p2;
33             p2 = p2->next;
34             if(p2==NULL&&carry==1)//一定要考虑这种进位了的情况
35             {
36                 pNew=malloc(sizeof(struct ListNode));
37                 pNew->val=1;
38                 pNew->next=NULL;
39                 pre->next=pNew;
40             }
41                 
42         }
43     }
44     else if (p1 != NULL)
45     {
46         while (p1)
47         {
48             res = p1->val + carry;
49             p1->val = res % 10;
50             carry = res / 10;
51             pre=p1;
52             p1 = p1->next;
53             if(p1==NULL&&carry==1)
54             {
55                 pNew=malloc(sizeof(struct ListNode));
56                 pNew->val=1;
57                 pNew->next=NULL;
58                 pre->next=pNew;
59             }
60         }
61     }
62     else
63     {
64         if (carry == 1)
65         {
66             pNew =malloc(sizeof(struct ListNode));
67             pNew->val = carry;
68             pNew->next = NULL;
69             pre->next = pNew;
70         }
71     }
72     return l1;
73 
74 }

手里拿着一把锤子，看什么都像钉子，编程界的锤子应该就是算法了吧！