【LeetCode刷题系列

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 = 0;
        // num2为两整数相加后的进位数值
        int num2 = 0;
        // node1Val为l1当前节点的数据域或者0
        int node1Val = 0;
        // node2Val为l2当前节点的数据域或者0
        int node2Val = 0;
        // 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 = 0;
        
        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 = 0;
            }else
            {
                node1Val = l1->val;
            }
            if(l2 == nullptr)
            {
                node2Val = 0;
            }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 >= 10)
            {
                num1 = num1 - 10;
                num2 = 1;
            }else
            {
                num2 = 0;
            }
            // 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 == 0)
            {
                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 != 0)
            {
                ListNode *newNode = new ListNode(num2);
                tail->next = newNode;
                tail = newNode;
                return head;
            }
            count++;
            
        }
    
        return head;
    }
};