Reverse doubly-link list in C++

Question

I've been trying to figure out how to reverse the order of a doubly-linked list, but for some reason, in my function void reverse() runs while loop once and then crashes for some reason. To answer some questions ahead, I'm self-teaching myself with my brothers help. This isn't all of the code, but I have a display() function which prints all nodes chronologically from start_ptr and a switch which activates certain functions like

    case 1 : add_end(); break;
    case 2 : add_begin(); break;
    case 3 : add_index(); break;
    case 4 : del_end(); break;
    case 5 : del_begin(); break;
    case 6 : reverse(); break;

This is the geist of my code:

#include <iostream>
using namespace std;

struct node
{
    char name[20];
    char profession[20];
    int age;
    node *nxt;
    node *prv;
};

node *start_ptr = NULL;

void pswap (node *pa, node *pb)
{
    node temp = *pa;
    *pa = *pb;
    *pb = temp;
    return;
}

void reverse()
{
    if(start_ptr==NULL)
    {
        cout << "Can't do anything" << endl;
    }
    else if(start_ptr->nxt==NULL)
    {
        return;
    }
    else
    {
        node *current = start_ptr;
        node *nextone = start_ptr;
        nextone=nextone->nxt->nxt;
        current=current->nxt;
        start_ptr->prv=start_ptr->nxt;
        start_ptr->nxt=NULL;
        //nextone=nextone->nxt;
        while(nextone->nxt!= NULL)
        {
            pswap(current->nxt, current->prv);
            current=nextone;
            nextone=nextone->nxt;
        }
        start_ptr=nextone;
    }
}

Answer 1

I thought I'd add a recursive solution here.

node* reverse_and_get_new_head(node* head) {
    if (head == nullptr) { return nullptr; } 
      // This can be avoided by ensuring the initial,
      // outer call is with a non-empty list
    std::swap(head->prev, head->next);
    if (head->prev == nullptr) { return head; }
    return reverse_and_get_new_head(head->prev);
}

void reverse() {
    start_ptr = reverse_and_get_new_head(start_ptr);
}

Answer 2

My code for reversing doubly linked list,

Node* Reverse(Node* head)
{
// Complete this function
// Do not write the main method. 


if(head != NULL) {

    Node* curr = head;
    Node* lastsetNode = curr;
    while(curr != NULL) {

        Node* frwdNode = curr->next;
        Node* prevNode = curr->prev;


        if(curr==head) {                
            curr->next = NULL;
            curr->prev = frwdNode;
            lastsetNode = curr;
        }
        else {
            curr->next = lastsetNode;
            curr->prev = frwdNode;
            lastsetNode = curr;
        }



        curr = frwdNode;
    }

    head = lastsetNode;
}


return head;
}

Answer 3

A very simple and O(n) solution using two pointers:

start = head of the doubly LL

struct node *temp, *s;
s = start;

while(s != NULL){
  temp = s->prev;
  s->prev = s->next;
  s->next = temp;
  s = s->prev;
}

//if list has more than one node
if(current != NULL){
  start = temp->prev;
}

Answer 4

Simple solution. reverses in less than half a number of total iterations over the list

template<typename E> void DLinkedList<E>::reverse() {
    int median = 0;
    int listSize = size();
    int counter = 0;

    if (listSize == 1)
    return;

    DNode<E>* tempNode = new DNode<E>();

    /**
     * A temporary node for swapping a node and its reflection node
     */
    DNode<E>* dummyNode = new DNode<E>();

    DNode<E>* headCursor = head;
    DNode<E>* tailCursor = tail;

    for (int i = 0; i < listSize / 2; i++) {
        cout << i << "\t";

        headCursor = headCursor->next;
        tailCursor = tailCursor->prev;

        DNode<E>* curNode = headCursor;
        DNode<E>* reflectionNode = tailCursor;

        if (listSize % 2 == 0 && listSize / 2 - 1 == i) {
            /**
             * insert a dummy node for reflection
         * for even sized lists
         */
        curNode->next = dummyNode;
        dummyNode->prev = curNode;

        reflectionNode->prev = dummyNode;
        dummyNode->next = reflectionNode;

    }
    /**
     * swap the connections from previous and 
             * next nodes for current and reflection nodes
     */

    curNode->prev->next = curNode->next->prev = reflectionNode;

    reflectionNode->prev->next = reflectionNode->next->prev = curNode;

    /**
     * swapping of the nodes
     */

    tempNode->prev = curNode->prev;
    tempNode->next = curNode->next;

    curNode->next = reflectionNode->next;
    curNode->prev = reflectionNode->prev;

    reflectionNode->prev = tempNode->prev;
    reflectionNode->next = tempNode->next;

    if (listSize % 2 == 0 && listSize / 2 - 1 == i) {
        /**
         * remove a dummy node for reflection
         * for even sized lists
         */
        reflectionNode->next = curNode;
        curNode->prev = reflectionNode;
    }

    /**
     * Reassign the cursors to position over the recently swapped nodes
     */
        tailCursor = curNode;
        headCursor = reflectionNode;

    }

    delete tempNode, dummyNode;
}

template<typename E> int DLinkedList<E>::size() {
    int count = 0;
    DNode<E>* iterator = head;

    while (iterator->next != tail) {
        count++;
        iterator = iterator->next;
    }
    return count;
}

Answer 5

You can simplify your reverse() quite a bit. I'd do something like this:

void reverse()
{
    if(start_ptr == NULL)
    {
        cout << "Can't do anything" << endl;
    }
    else
    {
        node *curr = start_ptr;
        while(curr != NULL)
        {
            Node *next = curr->next;
            curr->next = curr->prev;
            curr->prev = next;
            curr = next;
        }
        start_ptr = prev;       
    }
}

Explanation: The basic idea is simply to visit each Node and swap the links to previous and next. When we move curr to the next Node, we need to store the next node so we still have a pointer to it when we set curr.next to prev.

Answer 6

EDIT: My first implementation, which was correct but not perfect. Your implementation is pretty complicated. Can you try this instead:

node * reverse(Node * start_ptr)
{
    Node *curr = start_ptr; 
    Node * prev = null;
    Node * next = null;
    while(curr)
    {
        next = curr->nxt;
        curr->nxt = prev;
    curr->prv = next;
        prev = curr;
        curr = next;
    }
    return start_ptr=prev;
}

Here is my updated solution:

node * reverse()
{
    node *curr = start_ptr; 
    node * prev = NULL;
    node * next = NULL;
    while(curr)
    {
        next = curr->nxt;
        curr->nxt = prev;
        curr->prv = next;
        prev = curr;
        curr = next;
    }
    return start_ptr=prev;
}

The logic was correct. But the issue was that I was accepting in input argument start_ptr. Which means that I was returning the local copy of it. Now it should be working.

Answer 7

Your pswap function is wrong your should swap the pointer not try to create temporary objects and swap them. Should be like that (there might be other mistake later)

void pswap (node *&pa, node *&pb)
{
    node* temp = pa;
    pa = pb;
    pb = temp;
    return;
}

Answer 8

Look at

valuesnextone=nextone->nxt->nxt;

Here nextone->nxt can be null.

Apart from that, try to use pointers to pointers in the swap function.

Answer 9

I suggest maintaining a link to the last node.
If not, find the last node. Traverse the list using the "previous" links (or in your case, prv).

There is no need to actually change the links around. Traversing using the prv pointer will automatically visit the nodes in reverse order.

Answer 10

Try this:

node *ptr = start_ptr;
while (ptr != NULL) {
    node *tmp = ptr->nxt;
    ptr->nxt = ptr->prv;
    ptr->prv = tmp;
    if (tmp == NULL) {
        end_ptr = start_ptr;
        start_ptr = ptr;
    }
    ptr = tmp;
}