lintcode-【简单题】链表求和

题目：

你有两个用链表代表的整数，其中每个节点包含一个数字。数字存储按照在原来整数中相反的顺序，使得第一个数字位于链表的开头。写出一个函数将两个整数相加，用链表形式返回和。

样例：

给出两个链表 3->1->5->null 和 5->9->2->null，返回 8->0->8->null

答案：

从头到尾按链表顺序遍历相加就行啦，如果加到最后，进位不为0，还需要另外添加一个节点。

代码：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     /**
      * @param l1: the first list
      * @param l2: the second list
      * @return: the sum list of l1 and l2
      */
     ListNode *addLists(ListNode *l1, ListNode *l2) {
         // write your code here
         int carry = ,value = ;
         ListNode *l1Iter = l1;
         ListNode *l2Iter = l2;
         ListNode *ansRoot = NULL,*ansIter = NULL;

         while(l1Iter != NULL && l2Iter != NULL)
         {
             value = l1Iter->val + l2Iter->val + carry;
             carry = value / ;
             value = value % ;

             ListNode *node = new ListNode(value);
             if(ansRoot == NULL)
             {
                 ansRoot = node;
             }

             if(ansIter != NULL)
             {
                 ansIter->next = node;
             }

             ansIter = node;
             l1Iter = l1Iter->next;
             l2Iter = l2Iter->next;
         }

         while(l1Iter != NULL)
         {
             value = l1Iter->val + carry;
             carry = value / ;
             value = value % ;

             ListNode *node = new ListNode(value);
             if(ansRoot == NULL)
             {
                 ansRoot = node;
             }

             if(ansIter != NULL)
             {
                 ansIter->next = node;
             }

             ansIter = node;
             l1Iter = l1Iter->next;
         }

         while(l2Iter != NULL)
         {
             value = l2Iter->val + carry;
             carry = value / ;
             value = value % ;

             ListNode *node = new ListNode(value);
             if(ansRoot == NULL)
             {
                 ansRoot = node;
             }

             if(ansIter != NULL)
             {
                 ansIter->next = node;
             }

             ansIter = node;
             l2Iter = l2Iter->next;
         }

         if(carry != )
         {
             ListNode *node = new ListNode(carry);
             if(ansIter != NULL)
             {
                 ansIter->next = node;
             }
         }
         return ansRoot;
     }
 };