反向链接列表-C ++

Question

I wrote a function that should reverse a list.

So far, I can reverse only two items, but no more. I checked and double checked and still can't find the problem. I even used the Debugger to see the value of each pointer. When running the debugger, I received the message:

An access violation (segmentation fault) raised in your program.

This is my first assignment with linked lists so I am still learning.

Here is the code I wrote in Dev-C++:

List::ListNode *List::Reverse_List(ListNode *head)
{
    ListNode *cur = head;
    ListNode *forward = NULL;
    ListNode *previous = NULL;

    while (cur != NULL)
    {
        head = cur; //set the head to last node
        forward = head->next;  //save the next pointer in forward
        cur->next = previous;  //change next to previous
        previous = cur;
        cur = forward;

        cout << "cur= " << cur->item << endl; //this is just to display the current value of cur

        return head;
    }
}

Answer 1

Your code is close, it is returning early.

List::ListNode *List::Reverse_List(ListNode *head) 
{
    ListNode *cur = head;
    ListNode *forward = NULL;
    ListNode *previous = NULL;

    while (cur != NULL) {
        //There is no need to use head here, cur will suffice
        //head = cur; //set the head to last node
        forward = cur->next; //save the next pointer in forward

        cur->next = previous; //change next to previous
        previous = cur;
        cur = forward;

        cout << "cur= " << cur->item << endl; //this is just to display the current value of cur

        //don't return here you have only adjusted one node
        //return head;
    }

    //at this point cur is NULL, but previous still holds the correct node
    return previous;
}

Answer 2

Sorry for answering late and I am sure that you would have found the answer by now but yet it might be helpful for others. Answer is simply taking return statement (ie return head;) out of while loop will fix your problem. Though there are ways you can avoid extra pointer and assignments to optimise your code.

Answer 3

Everyone must have the same homework assignment today.

I think it would be more helpful to these people to show them what happens to the state of the list while it is being reversed. This should help them better than showing them code or code problems.

Here is what should happen (with the algorithm I would use)

[] = head () = current

([1])->2->3->4, [2]->(1)->3->4, [3]->2->(1)->4, [4]->3->2->(1) done because current now doesn't have a new next