将链表复制到对链表的引用

Question

I have been assigned to pass a pointer to a linked list and copy that to a new list passed as a pointer to a linked list, and copy it recursively. I receive a segmentation fault when trying to copy the very first item.

I've tried every combination of pointer and reference I can think of while maintaining the program requirements of the function prototype: void duplicate(node * head, node *& newHead)

 #include <iostream>
 #include "supplied.o"

 using namespace std;

 struct node
 {
     int data;
     node * next;
 };

 int main()
 {
     node * head = NULL;
     build(head);  // supplied function initializes list
     newHead = NULL;
     duplicate (head, newHead);
 }

 void duplicate (node * head, node*& newHead)
 {
     node * iterator1  = NULL;
     node * iterator2 = new Node;
     node * iterator2 = newHead;
     iterator2->data = iterator1->data; //error occurs here
     // program will continue to copy list recursively
 }

 void build (node *& head) //cannot see this function; provided via "supplied.o"
 {
 }

the error occurs because the function is unable to access iterator2->data. Iterator1->data can be accessed and even printed without problem.

Answer 1

Just create a new node, then copy the data, then recurse with the next nodes. Take care of null pointers. That's all.

duplicate(const node*head, node* &newhead)
{
    if(head) {
        // copy a valid node
        newhead = new node;                   // allocate a new node
        newhead->data = head->data;           // copy the data
        duplicate(head->next, newhead->next); // recurse on the next node
    } else
        // in case of null pointer: terminate recursion
        newhead = nullptr;
}

Now, consider what happens step-by-step if you call duplicate with a long list and convince yourself that it actually does what you want.