力扣刷题二——两数相加（链表逆序）+（注释详解）

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
 class Solution {
 public:
     ListNode* addTwoNumbers(ListNode* l1, ListNode* l2)
     {
         int len1 = 1; int len2 = 1;
         //让p1和p2代替头结点进行遍历
         ListNode* p1 = l1; ListNode* p2 = l2;

         //找出l1和l2的尾结点,记录l1和l2的长度
         while (p1->next != NULL)
         {
             p1 = p1->next;
             len1++;
         }

         while (p2->next != NULL)
         {
             p2 = p2->next;
             len2++;
         }
         //进行补0操作
         //假设len1比len2长
         if (len1 > len2)
         {
             for (int i = 1; i <= len1 - len2; i++)
             {
                 p2->next = new ListNode(0);
                 p2 = p2->next;
             }
         }
         //len2比len1长
         else
         {
             for (int i = 1; i <= len2 - len1; i++)
             {
                 p1->next = new ListNode(0);
                 p1 = p1->next;
             }
         }
         //进行相加操作
         p1=l1;
         p2=l2;
         int temp = 0;
         bool count=false;
         ListNode* l3=new ListNode(-1);//用于函数返回值的链表l3
         ListNode* w = l3;//代替l3的指针w
         int maxnum = len1 > len2 ? len1 : len2;
         for (int i = 1; i <= maxnum; i++)
         {
                 temp = count + p1->val + p2->val;
                 w->next = new ListNode(temp%10);//创建相加以后的链表
                 count = temp >= 10 ? true : false;
                 w = w->next;
                 p1 = p1->next;
                 p2 = p2->next;
         }
         //如果最后两个数相加还有进位，则再创建一个节点来存放
         if (count)
         {
             w->next = new ListNode(1);
             w = w->next;
         }
         return l3->next;

     }
 };