Reverse A Linked List in Group of a Given Size

Reverse a Linked List in groups of a given size

https://practice.geeksforgeeks.org/problems/reverse-a-linked-list-in-groups-of-given-size/1
Space complexity: O(0);
Time complexity: O(n);

 struct node *reverse(struct node *head, int k)
 {
  int n = k;
  if (head == 0 || head->next == 0) return head;
  struct  node *tempHead = head->next;
  head->next = 0;
  struct node *tail = head;// head will become tail later
         //reverse K element 
  while (tempHead != 0 && n>1) {
   struct node *temp = tempHead;
   tempHead = tempHead->next;
   temp->next = head;
   head = temp;
   n--;
  }
  tail->next = reverse(tempHead, k);
  return head;
 }







#include<iostream>
#include<string>

using namespace std;

class node {
public:
 int data;
 node *next;
 node(int d) {
  data = d;
  next = 0;
 }
};
class linkedList{
private:
 node *head;
public:
 linkedList() {
  head = 0;
 }
 void addFirst(int d) {
  node *newNode = new node(d);
  if (head == 0)
   head = newNode;
  else {
   newNode->next = head;
   head = newNode;
  }
 }
 void reverse(int k) {
  head=reverse(head, k);
 }
 struct node *reverse(struct node *head, int k)
 {
  int n = k;
  if (head == 0 || head->next == 0) return head;
  struct  node *tempHead = head->next;
  head->next = 0;
  struct node *tail = head;// head will become tail later
         //reverse K element 
  while (tempHead != 0 && n>1) {
   struct node *temp = tempHead;
   tempHead = tempHead->next;
   temp->next = head;
   head = temp;
   n--;
  }
  tail->next = reverse(tempHead, k);
  return head;
 }

  void print() {
   node *temp = head;
   while (temp != 0) {
    cout << temp->data << " ";
    temp = temp->next;
   }
   cout << endl;
  }
};
int main() {
 int n[] = { 1,2 ,3};
 //int n[] = { 1,2,2,4,5,6,7,8,9};
 linkedList a;
 for (int i = 2;i >=0;i--) {
  a.addFirst(n[i]);
 }
 a.print();
 a.reverse(4);
 a.print();
 system("PAUSE");
 return 0;

}

// by Y Nguyen