# Definition for singly-linked list.
#在节点ListNode定义中，定义为节点为结构变量。
#节点存储了两个变量：value 和 next。value 是这个节点的值，next 是指向下一节点的指针，当 next 为空指针时，这个节点是链表的最后一个节点。
#注意注意val只代表当前指针的值，比如p->val表示p指针的指向的值；而p->next表示链表下一个节点，也是一个指针。
#构造函数包含两个参数 _value 和 _next ，分别用来给节点赋值和指定下一节点
class ListNode:def __init__(self, val=0, next=None):self.val = valself.next = next
class Solution:def listnode_reversal(self,head:ListNode)->ListNode:print("开始反转")# head头节点，判断头节点是否为空，或者是否只存在一个节点if (head == None or head.next == None):return head;# 创建一个节点d = ListNode(-1)# 对于头节点来说相当于前面加了一个空的节点d.next = head# 定义前一节点指针，当前节点指针，辅助指针（后一节点指针）# pre节点一直都没有变pre = dcur = head# 让链表结构反转while (cur != None and cur.next != None):tmp = cur.next  # 用于保存当前节点指针cur的后一节点指针cur.next = tmp.next  # 更新cur指针tmp.next = pre.next  # 将当前节点指针cur后一节点的指针指向cur（即指针指向顺序颠倒）pre.next = tmp  # 更新pre指针return d.nextdef listnode_to_number(self,head)->int:if (head == None or head.next == None):return 0;# 计数sum = 1#数字resultnumber = 0while head != None:resultnumber = resultnumber + head.val * (10 ** sum)print("相加%d" % (resultnumber))sum = sum + 1head = head.nextreturn resultnumberdef addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:#反转列表l1 = self.listnode_reversal(l1)l2 = self.listnode_reversal(l2)#返回列表组成的数字resultl1 = self.listnode_to_number(l1)resultl2 = self.listnode_to_number(l2)print("l1倒叙组成的数字为%d"%(resultl1))print("l2倒叙组成的数字为%d" % (resultl2))resultsum = resultl1 + resultl2print("相加数字为%d" % (resultsum))#存放结果类型resultlistnodehead = ListNode(0)first_pointer = resultlistnodehead#结果的字符串类型strresult = str(resultsum)for k in strresult[::-1]:newnode = ListNode(int(k))resultlistnodehead.next = newnoderesultlistnodehead = newnodeprint(first_pointer.next.val)print(first_pointer.next.next.val)print(first_pointer.next.next.next.val)print(first_pointer.next.next.next.next.val)print(first_pointer.next.next.next.next.next.val)return first_pointer.nextl1 = ListNode(0);
l1_1 = ListNode(1);
l1.next = l1_1;
l1_2 = ListNode(2);
l1_1.next = l1_2;
l1_3 = ListNode(3);
l1_2.next = l1_3;
l1_4 = ListNode(4);
l1_3.next = l1_4;l2 = ListNode(0);
l2_1 = ListNode(1);
l2.next = l2_1;
l2_2 = ListNode(2);
l2_1.next = l2_2;
l2_3 = ListNode(3);
l2_2.next = l2_3;
l2_4 = ListNode(4);
l2_3.next = l2_4;
S = Solution()
print(S.addTwoNumbers(l1,l2))