Reverse Linked List II in C++

Suppose we have a linked list. We have to reverse the nodes from position m to n. We have to do it in one pass. So if the list is [1,2,3,4,5] and m = 2 and n = 4, then the result will be [1,4,,3,2,5]

Let us see the steps −

There will be two methods, the reverseN() and reverseBetween(). The reverseBetween() will work as main method.
define one link node pointer called successor as null
The reverseN will work as follows −
if n = 1, then successor := next of head, and return head
last = reverseN(next of head, n - 1)
next of (next of head) = head, and next of head := successor, return last
the reverseBetween() method will be like −
if m = 1, then return reverseN(head, n)
next of head := reverseBetween(next of head, m – 1, n – 1)

Let us see the following implementation to get better understanding −

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class ListNode{
   public:
      int val;
      ListNode *next;
      ListNode(int data){
         val = data;
         next = NULL;
      }
   };
   ListNode *make_list(vector<int> v){
      ListNode *head = new ListNode(v[0]);
      for(int i = 1; i<v.size(); i++){
         ListNode *ptr = head;
         while(ptr->next != NULL){
            ptr = ptr->next;
         }
         ptr->next = new ListNode(v[i]);
      }
      return head;
   }
   void print_list(ListNode *head){
      ListNode *ptr = head;
      cout << "[";
      while(ptr){
         cout << ptr->val << ", ";
         ptr = ptr->next;
      }
      cout << "]" << endl;
 }
 class Solution {
   public:
      ListNode* successor = NULL;
      ListNode* reverseN(ListNode* head, int n ){
         if(n == 1){
            successor = head->next;
            return head;
         }
         ListNode* last = reverseN(head->next, n - 1);
         head->next->next = head;
         head->next = successor;
         return last;
      }
      ListNode* reverseBetween(ListNode* head, int m, int n) {
      if(m == 1){
            return reverseN(head, n);
      }
      head->next = reverseBetween(head->next, m - 1, n - 1);
            return head;
   }
 };
main(){
   Solution ob;
   vector<int> v = {1,2,3,4,5,6,7,8};
   ListNode *head = make_list(v);
   print_list(ob.reverseBetween(head, 2, 6));
}