【LeetCode刷题系列 - 002题】Add Two Numbers

题目：

You are given two non-empty linked lists representing two non-negative integers. 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.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

代码(C++实现)：

 /**
  * Definition for singly-linked list.
  * struct ListNode {
  *     int val;
  *     ListNode *next;
  *     ListNode(int x) : val(x), next(NULL) {}
  * };
  */
 class Solution {
 public:
     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
         // 结果链表的首节点
         ListNode *head = nullptr;
         // 尾插法建立单链表指向尾节点
         ListNode *tail = nullptr;
         // num1为两整数相加后的个位数值
         int num1 = ;
         // num2为两整数相加后的进位数值
         int num2 = ;
         // node1Val为l1当前节点的数据域或者0
         int node1Val = ;
         // node2Val为l2当前节点的数据域或者0
         int node2Val = ;
         // count变量的设置是为了尾插法建立单链表设置的计数器
         int count = ;

         while(l1 != nullptr || l2 != nullptr)
         {
             // 如果节点不为空，则取节点的数据域否则让节点的数据域为0
             if(l1 == nullptr)
             {
                 node1Val = ;
             }else
             {
                 node1Val = l1->val;
             }
             if(l2 == nullptr)
             {
                 node2Val = ;
             }else
             {
                  node2Val = l2->val;
             }
             // 本次计算结果 = 本次计算的node1Val + 本次计算的node1Va2 + 进位值
             num1 = node1Val + node2Val + num2;
             if(num1 >= )
             {
                 num1 = num1 - ;
                 num2 = ;
             }else
             {
                 num2 = ;
             }
             // 为建立结果链表创建节点
             ListNode *newNode = new ListNode(num1);
             // 尾插法建立结果单链表，如果是首节点，需要进行特殊处理
             if(count == )
             {
                 head = tail = newNode;
             }else
             {
                 tail->next = newNode;
                 tail = newNode;
             }
             // 链表向前移动并释放原始链表所占的内存空间
             if(l1 != nullptr)
             {
                 ListNode *tempNode1 = l1;
                 l1 = l1->next;
                 delete tempNode1;
             }
             if(l2 != nullptr)
             {
                 ListNode *tempNode2 = l2;
                 l2 = l2->next;
                 delete tempNode2;
             }

             /* 为了解决例如情况：l1 = [5]
                               l2 = [5]
                这种情况
             */
             if(l1 == nullptr && l2 == nullptr && num2 != )
             {
                 ListNode *newNode = new ListNode(num2);
                 tail->next = newNode;
                 tail = newNode;
                 return head;
             }
             count++;

         }

         return head;
     }
 };