链表-Add Two Numbers

第一版代码（很挫很罗嗦，不过是第一次做，记录一下成长的脚步！继续努力！）

 /*struct ListNode {
      int val;
      struct ListNode *next;
 };*/

 typedef struct ListNode ListNode;

 struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) {
     ListNode *result = (ListNode *)malloc(sizeof(ListNode));
     result->next = NULL;
     ListNode *now = result;

     ListNode *p1 = l1;
     ListNode *p2 = l2;
     int carry = ;

     while(p1&&p2)
     {
         ListNode *p = (ListNode *)malloc(sizeof(ListNode));
         p->next = NULL;
         int r = p1->val + p2->val + carry;
         carry = r/;
         p->val = r%;
         now->next = p;
         now = now->next;
         p1 = p1->next;
         p2 = p2->next;
     }

     while(p1)
     {
         ListNode *p = (ListNode *)malloc(sizeof(ListNode));
         p->next = NULL;
         int r = p1->val + carry;
         carry = r/;
         p->val = r%;
         now->next = p;
         now = now->next;
         p1 = p1->next;
     }

     while(p2)
     {
         ListNode *p = (ListNode *)malloc(sizeof(ListNode));
         p->next = NULL;
         int r = p2->val + carry;
         carry = r/;
         p->val = r%;
         now->next = p;
         now = now->next;
         p2 = p2->next;
     }
     if(carry)
     {
         ListNode *p = (ListNode *)malloc(sizeof(ListNode));
         p->next = NULL;
         p->val = carry;
         now->next = p;
         now = now->next;
     }

     return result->next;

 }

要注意返回的内容必须符合题目要求，一开始我直接返回了result，但是leetcode测试的时候会把空的头结点也输出，不符合题目要求。

要全面考虑问题，这题的特殊情况主要体现在进位部分，如999+1这种情况，还有就是两个数的位数可能会不同。

可以使用条件操作符对代码进行简化，简化版如下：

 void addTwoNumbers(struct ListNode* l1, struct ListNode* l2)
 {
     ListNode *result = (ListNode *)malloc(sizeof(ListNode));
     result->next = NULL;
     ListNode *now = result;

     ListNode *p1 = l1;
     ListNode *p2 = l2;
     int carry = ;

     while(p1&&p2)
     {
         ListNode *p = (ListNode *)malloc(sizeof(ListNode));
         p->next = NULL;
         int r = p1->val + p2->val + carry;
         carry = r/;   //进位
         p->val = r%;
         now->next = p;  //插入到result链表中
         now = now->next;
         p1 = p1->next;
         p2 = p2->next;
     }

     ListNode *p3 = p1?p1:p2;
     while(p3)
     {
         ListNode *p = (ListNode *)malloc(sizeof(ListNode));
         p->next = NULL;
         int r = p3->val + carry;
         carry = r/;   //进位
         p->val = r%;
         now->next = p;  //插入到result链表中
         now = now->next;
         p3 = p3->next;
     }

     if(carry)
     {
         ListNode *p = (ListNode *)malloc(sizeof(ListNode));
         p->next = NULL;
         p->val = carry;
         now->next = p;  //插入到result链表中
         now = now->next;
     }

     Print(result->next);

 }

【待补充C++版本。。。】