【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 -> 8Explanation: 342   465 = 807.

代码(C 实现)：

 1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {12         // 结果链表的首节点13         ListNode *head = nullptr; 14         // 尾插法建立单链表指向尾节点15         ListNode *tail = nullptr; 16         // num1为两整数相加后的个位数值17         int num1 = 0;18         // num2为两整数相加后的进位数值19         int num2 = 0;20         // node1Val为l1当前节点的数据域或者021         int node1Val = 0;22         // node2Val为l2当前节点的数据域或者023         int node2Val = 0;24         // count变量的设置是为了尾插法建立单链表设置的计数器25         int count = 0;26         27         while(l1 != nullptr || l2 != nullptr)28         {29             // 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030             if(l1 == nullptr)31             {32                 node1Val = 0;33             }else34             {35                 node1Val = l1->val;36             }37             if(l2 == nullptr)38             {39                 node2Val = 0;40             }else41             {42                  node2Val = l2->val;43             }44             // 本次计算结果 = 本次计算的node1Val   本次计算的node1Va2   进位值45             num1 = node1Val   node2Val   num2;46             if(num1 >= 10)47             {48                 num1 = num1 - 10;49                 num2 = 1;50             }else51             {52                 num2 = 0;53             }54             // 为建立结果链表创建节点55             ListNode *newNode = new ListNode(num1);56             // 尾插法建立结果单链表，如果是首节点，需要进行特殊处理57             if(count == 0)58             {59                 head = tail = newNode;60             }else61             {62                 tail->next = newNode;63                 tail = newNode;64             }65             // 链表向前移动并释放原始链表所占的内存空间66             if(l1 != nullptr)67             {68                 ListNode *tempNode1 = l1;69                 l1 = l1->next;70                 delete tempNode1;71             }  72             if(l2 != nullptr)73             {74                 ListNode *tempNode2 = l2;75                 l2 = l2->next;76                 delete tempNode2;77             }78             79             /* 为了解决例如情况：l1 = [5]  80                               l2 = [5] 81                这种情况82             */83             if(l1 == nullptr && l2 == nullptr && num2 != 0)84             {85                 ListNode *newNode = new ListNode(num2);86                 tail->next = newNode;87                 tail = newNode;88                 return head;89             }90             count  ;91             92         }93     94         return head;95     }96 };

 

来源：http://www.icode9.com/content-4-173401.html