Reverse Nodes in k-Group in C++

C++Server Side ProgrammingProgramming

Suppose we have a linked list, we have to reverse the nodes of the linked list k at a time and return its modified list. Here k is a positive integer and is less than or equal to the length of the linked list. So if the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

So if the linked list is like [1,2,3,4,5,6,7] and k is 3, then the result will be [3,2,1,6,5,4,7].

To solve this, we will follow these steps −

  • Define a method called solve(), this will take the head of the linked list, take partCount and k

  • if partCount is 0, then return head

  • newHead := head, prev := null, x := k

  • while newHead is not null and x is not 0

    • temp := next of newHead, next of nextHead := prev

    • prev := newHead, newHead := temp

  • next of head := solve(newHead, partCount – 1, k)

  • return prev

  • From the main method do the following −

  • return solve(head of the linked list, length of list / k, k)

Example

Let us see the following implementation to get a better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
void print_vector(vector<auto> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << v[i] << ", ";
   }
   cout << "]"<<endl;
}
void print_vector(vector<vector<auto>> v){
   cout << "[";
   for(int i = 0; i<v.size(); i++){
      cout << "[";
      for(int j = 0; j <v[i].size(); j++){
         cout << v[i][j] << ", ";
      }
      cout << "],";
   }
   cout << "]"<<endl;
}
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* solve(ListNode* head, int partitionCount, int k){
      if(partitionCount == 0)return head;
      ListNode *newHead = head;
      ListNode* prev = NULL;
      ListNode* temp;
      int x = k;
      while(newHead && x--){
         temp = newHead->next;
         newHead->next = prev;
         prev = newHead;
         newHead = temp;
      }
      head->next = solve(newHead, partitionCount - 1, k);
      return prev;
   }
   int calcLength(ListNode* head){
      int len = 0;
      ListNode* curr = head;
      while(curr){
         len++;
         curr = curr->next;
      }
      return len;
   }
   ListNode* reverseKGroup(ListNode* head, int k) {
      int length = calcLength(head);
      return solve(head, length / k, k);
   }
};
main(){
   vector<int> v = {1,2,3,4,5,6,7};
   ListNode *head = make_list(v);
   Solution ob;
   print_list(ob.reverseKGroup(head, 3));
}

Input

1,2,3,4,5,6,7
3

Output

[3, 2, 1, 6, 5, 4, 7, ]
raja
Published on 26-May-2020 11:32:45
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