Reverse Linked List II in C++\\n

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, ]