思路：先对两个链表进行反转，反转求和注意进位运算，求和完成之后再进行反转得到结果
/*** struct ListNode {* int val;* struct ListNode *next;* ListNode(int x) : val(x), next(nullptr) {}* };*/
#include <cstddef>
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 hereListNode* phead1 = ReverseList(head1);ListNode* phead2 = ReverseList(head2);int i= 0;ListNode* result = new ListNode(0);ListNode* ret = result;while(phead2||phead1||i>0){int val1 = phead1?phead1->val:0;int val2 = phead2?phead2->val:0;int sum = val1 + val2 + i;i = sum/10;sum = sum%10;ret->next = new ListNode(sum);ret = ret->next;if(phead1) phead1 = phead1->next;if(phead2) phead2 = phead2->next;}return ReverseList(result->next);}ListNode* ReverseList(ListNode* head){ListNode* ret = nullptr;ListNode* temp = nullptr;while(head){temp = head->next;head->next = ret;ret = head;head = temp;}return ret;}};
