Rotate a Linked List by K Places in C++

Suppose we have a linked list. We have to rotate the list to the right by k places. The value of k will be positive. So if the list is like [1 −> 2 −> 3 −> 4 −> 5 −> NULL], and k = 2, then the output will be [4 −> 5 −> 1 −> 2 −> 3 −> NULL]

Let us see the steps −

• If the list is empty, then return null

• len := 1

• create one node called tail := head

• while next of tail is not null

  • increase len by 1

  • tail := next of tail

• next of tail := head

• k := k mod len

• newHead := null

• for i := 0 to len − k

  • tail := next of tail

• newHead := next of tail

• next of tail := null

• return newHead

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->next){
      cout << ptr->val << ", ";
      ptr = ptr->next;
   }
   cout << "]" << endl;
}
class Solution {
   public:
   ListNode* rotateRight(ListNode* head, int k) {
      if(!head) return head;
      int len = 1;
      ListNode* tail = head;
      while(tail->next){
         len++;
         tail = tail->next;
      }
      tail->next = head;
      k %= len;
      ListNode* newHead = NULL;
      for(int i = 0; i < len - k; i++){
         tail = tail->next;
      }
      newHead = tail->next;
      tail->next = NULL;
      return newHead;
   }
};
main(){
   Solution ob;
   vector<int> v = {1,2,3,4,5,6,7,8,9};
   ListNode *head = make_list(v);
   print_list(ob.rotateRight(head, 4));
}

Input

[1,2,3,4,5,6,7,8,9], 4

Output

[6, 7, 8, 9, 1, 2, 3, 4, ]