Reverse Linked List II in C++

C++Server Side ProgrammingProgramming

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));
}

Input

[1,2,3,4,5,6,7,8]
2
6

Output

[1, 6, 5, 4, 3, 2, 7, 8, ]
raja
Published on 03-Feb-2020 15:00:08
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