1. LeetCode 206 反转链表

题目链接🔗

解题思路： 🔍
🐧创建一个新的节点，使用链表头插的方法；

2. NC40 链表相加

题目链接🔗

解题思路： 🔍
🐧 将两个链表进行逆置，然后再创建一个结果集的链表，利用头插，将结果插入；

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param head1 ListNode类 
     * @param head2 ListNode类 
     * @return ListNode类
     */
    ListNode* addInList(ListNode* head1, ListNode* head2) {
        // write code here
        ListNode* newHead1 = new ListNode(0);


        // 先将链表逆置
        ListNode* cur1 = head1;
        while (cur1 != nullptr)
        {
            ListNode* tmp = cur1->next;

            cur1->next = newHead1->next;

            newHead1->next = cur1;

            cur1 = tmp;
        }

        // 将链表逆置
        ListNode* newHead2 = new ListNode(0);
        ListNode* cur2 = head2;
        while (cur2 != nullptr)
        {
            ListNode* tmp = cur2->next;

            cur2->next = newHead2->next;

            newHead2->next = cur2;

            cur2 = tmp;
        }

        ListNode* resHead = new ListNode(0);
        cur1 = newHead1->next;
        cur2 = newHead2->next;

        int k = 0;
        while (cur1 || cur2)
        {
            int num1 = 0;
            if (cur1 != nullptr)
                num1 = cur1->val;
            
            int num2 = 0;
            if (cur2 != nullptr)
                num2 = cur2->val;

            // 每次把结果头插
            int num = num1 + num2 + k;

            if (num >= 10)
                k = 1;
            else
                k = 0;

            num %= 10;

            ListNode* numNode = new ListNode(num);

            numNode->next = resHead->next;
            resHead->next = numNode;

            if (cur1 != nullptr)
                cur1 = cur1->next;

            if (cur2 != nullptr)
                cur2 = cur2->next;
        }   

        if (k > 0)
        {
            ListNode* numNode = new ListNode(1);

            numNode->next = resHead->next;
            resHead->next = numNode;
        }
        return resHead->next;
    }