双链表相加问题

You are given two linked lists representing two non-negative numbers. e 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

此算法虽然不具备什么难度，但是如果从抽象的推演到可运行的代码，其中有不少的细节需要注意。其中涉及到了进位问题，两个链表长度也可能不同，实现代码如下所示：

#include <iostream>

struct ListNode
{
  int    val;
  ListNode*  next;

  ListNode(int _val)
  {
    val = _val;
  }
};

class Solution
{
public:
  ListNode* addTwoNumber(ListNode* l1, ListNode* l2)
  {
    static ListNode dummy(-1);
    int carry = 0;
    ListNode* prev = &dummy;

    for(ListNode *pa=l1, *pb=l2;
      pa != NULL || pb!= NULL;
      pa = pa==NULL?NULL:pa->next,
      pb = pb==NULL?NULL:pb->next,
      prev = prev->next)
    {
      const int ai = pa==NULL?0:pa->val;
      const int bi = pb==NULL?0:pb->val;

      const int value = (ai+bi+carry) % 10;
      carry = (ai+bi+carry)/10;

      prev->next = new ListNode(value);
    }

    if ( carry > 0)
    {
      prev->next = new ListNode(carry);
      
    }

    prev->next->next = NULL;
    return dummy.next;
  }

  void show(ListNode* ln)
  {
    for(; ln!= NULL; ln=ln->next)
    {
      printf("%d\n", ln->val);
    }
  }
};

int main()
{
  ListNode l1(2), l2(3), l3(7);
  l1.next = &l2;
  l2.next = &l3;
  l3.next = NULL;

  ListNode k1(8), k2(5), k3(6);
  k1.next = &k2;
  k2.next = &k3;
  k3.next = NULL;

  Solution so;
  ListNode* ad = so.addTwoNumber(&l1, &k1);
  so.show(ad);
  return 0;
}