Add Two Numbers

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

问题描述：给定两个链表，每个链表当中代表两个非负数字，根据输入和输出可以看到

输入：   2   4   3

         5   6   4

输出：   7   0   8

可以看出根据链表依次进行相加小于10，则直接添加在新建列表中，如果大于10，则存储进位值，作为下一节点的值进行添加

class Solution {
private:
  ListNode* ret;
public:
  ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
    ListNode* head_1 = l1,*head_2=l2;
    int index = 0;
    int value = 0;
    int high_value = 0;
     ret= new ListNode(0);
    ListNode* last_ret = ret;
    while (head_1 != NULL&&head_2 != NULL)
    {
      value = head_1->val + head_2->val+high_value;
      if (value < 10)
      {
        high_value = 0;
        last_ret->val = value;
      }
      else
      {
        high_value = value / 10;
        value = value % 10;
        last_ret->val = value;
      }
      head_1 = head_1->next;
      head_2 = head_2->next;
      if (head_1 != NULL&&head_2 != NULL)
      {
        last_ret->next = new ListNode(0);
        last_ret = last_ret->next;
      }
    }
    ListNode* cur = NULL;
    if (head_1 != NULL)
    {
      cur = head_1;
    }
    else
    {
      cur = head_2;
    }
    if (cur != NULL)
    {

      last_ret->next = new ListNode(0);
      last_ret = last_ret->next;
      while (cur != NULL)
      {
        value = cur->val + high_value;
        if (value < 10)
        {
          high_value = 0;
        }
        else
        {
          high_value = value / 10;
          value = value % 10;
        }
        last_ret->val = value;
        cur = cur->next;
        if (cur != NULL)
        {
          last_ret->next = new ListNode(0);
          last_ret = last_ret->next;
        }

      }     
    }
    if (high_value > 0)
    {
      last_ret->next = new ListNode(0);
      last_ret = last_ret->next;
      last_ret->val = high_value;
    }
    return ret;
  }
  ~Solution()
  {
    ListNode* cur = NULL;
    cur = ret;
    while (cur != NULL)
    {
      ListNode* m = cur;
      if (cur != NULL)
      {
        cur = cur->next;
      }
      if(m)
      delete m;
    }
  }
};