leetcode[2]:两数相加 C语言解法

给出两个 非空 的链表用来表示两个非负的整数。其中，它们各自的位数是按照 逆序 的方式存储的，并且它们的每个节点只能存储 一位 数字。

如果，我们将这两个数相加起来，则会返回一个新的链表来表示它们的和。

您可以假设除了数字 0 之外，这两个数都不会以 0 开头。

示例：

输入：(2 -> 4 -> 3) + (5 -> 6 -> 4)
输出：7 -> 0 -> 8
原因：342 + 465 = 807

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
struct ListNode* addTwoNumbers(struct ListNode* l1, struct ListNode* l2) 
{
    int remainder   = 0;
    int integer     = 0;
    int sum         = 0;
    int l1_val      = 0;
    int l2_val      = 0;
    struct ListNode *l_end = (struct ListNode *)malloc(sizeof(struct ListNode));
    struct ListNode *l_head = (struct ListNode *)malloc(sizeof(struct ListNode));
    struct ListNode *l_node;
    struct ListNode *l1_p = (struct ListNode *)malloc(sizeof(struct ListNode));
    struct ListNode *l2_p = (struct ListNode *)malloc(sizeof(struct ListNode));
    
    l1_p = l1;
    l2_p = l2;

    /*尾插法，当前只有头结点，且为空*/
    l_head->next = NULL;
    l_end = l_head;
    
    while((l1_p != NULL) || (l2_p != NULL))
    {

        l1_val = (l1_p != NULL)?l1_p->val:0;
        l2_val = (l2_p != NULL)?l2_p->val:0;
        sum       = l1_val + l2_val + integer;
        remainder = sum %10;
   
        
        l_node = (struct ListNode *)malloc(sizeof(struct ListNode));
        l_node->next = NULL;
        l_node->val = remainder;
        l_end->next = l_node;
        l_end = l_node;
        
        if(l1_p != NULL)
        {
            l1_p = l1_p->next;
        }
        
        if(l2_p != NULL)
        {
            l2_p = l2_p->next;
        }

        integer   = sum /10;
  
    }
    
     if(integer > 0)
     {
        l_node = (struct ListNode *)malloc(sizeof(struct ListNode));
        l_node->next = NULL;
        l_node->val = integer;
        l_end->next = l_node;
        l_end = l_node;
     }
    
    return l_head->next;
    
}