You have two numbers represented by a linked list, where each node contains a single digit The digits are stored in reverse order, such that the 1’s digit is at the head of the list. Write a function that adds the two numbers and returns the sum as a linked list.

EXAMPLE Input: (3 -> 1 -> 5) + (5 -> 9 -> 2)

Output: 8 -> 0 -> 8

我采用非递归，按照答案的思路写了递归的方法

#include 
    
     using namespace std;// Creating a Linked List:struct Node{    Node(int d):data(d){next = NULL;};    int data;    Node* next;};void printNode(Node* head){    cout<<"Print:"<
     
      data<
      
       next;    }    cout<<"End"<
       
        data) + (o2==NULL?0:o2->data) + flag;        flag = temp / 10;        q = new Node(temp % 10);        p->next = q;        p = p->next;        if(o1 != NULL)            o1 = o1->next;        if(o2 != NULL)            o2 = o2->next;    }    p->next = NULL;    p = head->next;    delete head;    return p;  };Node* plusNodeRec(Node *o1, Node *o2, int carry){    if(o1 == NULL && o2 == NULL)        return NULL;    int result = (o1==NULL?0:o1->data) + (o2==NULL?0:o2->data) + carry;    Node* p = new Node(result % 10);    Node* next = plusNodeRec(o1==NULL?NULL:o1->next,             o2==NULL?NULL:o2->next, result / 10);    p->next = next;    return p;};int main(){    Node* head1 = new Node(3);    head1->next = new Node(1);    Node* head2 = new Node(5);    head2->next = new Node(9);    head2->next->next = new Node(2);    // printNode(plusNode(head1, head2));    printNode(plusNodeRec(head1, head2, 0));    system("pause"); };