C++ - Reverse the Doubly Linked List

While working with a Doubly Linked List, sometimes it is required to reverse it. Reversing a List produces following result: if the given List is 10->20->30->40->50, after reversing the List the List becomes 50->40->30->20->10.

Reversing a List requires creating three nodes, considering that the list is not empty, which are as follows: tempNode pointing to head, prevNode pointing to head and curNode pointing to next of head. Then make next and previous of the prevNode as null to make the first node as last node of the reversed list. After that, while the curNode is not null adjust links (unlink curNode and link it to the reversed list from front and modify curNode and prevNode to the next element in the list). At last, make the prevNode (last node) as head.

The function reverseList is created for this purpose. It is a 4-step process.

void reverseList() {
  //1. If head is not null create three nodes
  //   prevNode - pointing to head,
  //   tempNode - pointing to head,
  //   curNode - pointing to next of head
  if(head != NULL) {
    Node* prevNode = head;
    Node* tempNode = head;
    Node* curNode = head->next;
    
    //2. assign next and previous of the 
    //   prevNode as null to make the first 
    //   node as last node of the reversed list
    prevNode->next = NULL;
    prevNode->prev = NULL;

    while(curNode != NULL) {
      //3. While the curNode is not null adjust links 
      //   (unlink curNode and link it to the reversed list 
      //   from front and modify curNode and prevNode) 
      tempNode = curNode->next;
      curNode->next = prevNode;
      prevNode->prev = curNode;
      prevNode = curNode;
      curNode = tempNode;
    }

    //4. Make prevNode (last node) as head
    head = prevNode;
  }
}

The below is a complete program that uses above discussed concept to reverse a given doubly linked list.

#include <iostream>
using namespace std;

//node structure
struct Node {
    int data;
    Node* next;
    Node* prev;
};

class LinkedList {
  private:
    Node* head;
  public:
    LinkedList(){
      head = NULL;
    }
 
    //Add new element at the end of the list
    void push_back(int newElement) {
      Node* newNode = new Node();
      newNode->data = newElement;
      newNode->next = NULL;
      newNode->prev = NULL; 
      if(head == NULL) {
        head = newNode;
      } else {
        Node* temp = head;
        while(temp->next != NULL)
          temp = temp->next;
        temp->next = newNode;
        newNode->prev = temp;
      }    
    }

    //reverse the list
    void reverseList() {
      if(head != NULL) {
        Node* prevNode = head;
        Node* tempNode = head;
        Node* curNode = head->next;
        
        prevNode->next = NULL;
        prevNode->prev = NULL;

        while(curNode != NULL) {
          tempNode = curNode->next;
          curNode->next = prevNode;
          prevNode->prev = curNode;
          prevNode = curNode;
          curNode = tempNode;
        }

        head = prevNode;
      }
    } 

    //display the content of the list
    void PrintList() {
      Node* temp = head;
      if(temp != NULL) {
        cout<<"The list contains: ";
        while(temp != NULL) {
          cout<<temp->data<<" ";
          temp = temp->next;
        }
        cout<<"\n";
      } else {
        cout<<"The list is empty.\n";
      }
    }   
};

// test the code 
int main() {
  LinkedList MyList;

  //Add five elements at the end of the list.
  MyList.push_back(10);
  MyList.push_back(20);
  MyList.push_back(30);
  MyList.push_back(40);
  MyList.push_back(50);

  //Display the content of the list.
  MyList.PrintList();
  
  //Reversing the list.
  MyList.reverseList();
  
  //Display the content of the list.
  MyList.PrintList();
  
  return 0; 
}

The above code will give the following output:

The list contains: 10 20 30 40 50 
The list contains: 50 40 30 20 10