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Reverse given Linked List in groups of specific given sizes
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Reverse a Linked List in groups of given size using Stack

Last Updated : 12 Sep, 2024
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Given a Singly linked list containing n nodes. The task is to reverse every group of k nodes in the list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should be considered as a group and must be reversed.

Examples: 

Input: head: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL, k = 2 
Output: head: 2 -> 1 -> 4 -> 3 -> 6 -> 5 -> NULL 
Explanation : Linked List is reversed in a group of size k = 2.

Reverse-a-Linked-List-in-groups-of-given-size-1


Input: head: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL, k = 4 
Output: head:  4 -> 3 -> 2 -> 1 -> 6 -> 5 -> NULL
Explanation : Linked List is reversed in a group of size k = 4.

Reverse-a-Linked-List-in-groups-of-given-size-2

Approach:

The idea is to use a Stack to store the nodes of the given linked list. Firstly, push the k nodes of the linked list into the stack. Now, pop the nodes one by one and keep track of the previously popped node. Point the next pointer of the prev node to the top element of the stack. Repeat this process, until we reach end of linked list.

Below is the implementation of the above approach: 

C++ 
// C++ program to reverse a linked list in groups // of given size  #include <bits/stdc++.h> using namespace std;  class Node {   public:     int data;     Node *next;      Node(int x) {         data = x;         next = nullptr;     } };  // Function to reverse the linked list in groups Node *reverseKGroup(Node *head, int k) {     if (!head || k == 1) {         return head;     }      stack<Node*> st;      Node *curr = head;      Node *prev = nullptr;         while (curr != nullptr) {             // Terminate the loop when either       	// current == NULL or count >= k          int count = 0;          while (curr != nullptr && count < k) {              st.push(curr);              curr = curr->next;              count++;          }             // Now pop the elements from the stack one by one          while (!st.empty()) {                           // If the final list has not been started yet             if (prev == nullptr) {                  prev = st.top();                  head = prev;                  st.pop();              } else {                  prev->next = st.top();                  prev = prev->next;                  st.pop();              }          }      }         // Set the next pointer of the    	// last node to NULL     prev->next = nullptr;         return head;  }  void printList(Node *head) {     Node *curr = head;     while (curr != nullptr) {         cout << curr->data << " ";         curr = curr->next;     }     cout << endl; }  int main() {          // Creating a sample singly linked list:     // 1 -> 2 -> 3 -> 4 -> 5 -> 6     Node *head = new Node(1);     head->next = new Node(2);     head->next->next = new Node(3);     head->next->next->next = new Node(4);     head->next->next->next->next = new Node(5);     head->next->next->next->next->next = new Node(6);      head = reverseKGroup(head, 3);          printList(head);      return 0; }
Java 
// Java program to reverse a linked list // in groups of given size  import java.util.Stack;  class Node {     int data;     Node next;      Node(int data) {         this.data = data;         this.next = null;     } }  // Function to reverse the linked list in groups class GfG {     static Node reverseKGroup(Node head, int k) {         if (head == null || k == 1) {             return head;         }          Stack<Node> st = new Stack<>();         Node curr = head;         Node prev = null;          while (curr != null) {              // Terminate the loop when either            	// current == null or count >= k             int count = 0;             while (curr != null && count < k) {                 st.push(curr);                 curr = curr.next;                 count++;             }              // Now pop the elements from the stack one by one             while (!st.isEmpty()) {                                // If the final list has not been started yet                 if (prev == null) {                     prev = st.pop();                     head = prev;                 } else {                     prev.next = st.pop();                     prev = prev.next;                 }             }         }          // Set the next pointer of the last node to null         prev.next = null;          return head;     }      static void printList(Node head) {         Node curr = head;         while (curr != null) {             System.out.print(curr.data + " ");             curr = curr.next;         }         System.out.println();     }      public static void main(String[] args) {                // Creating a sample singly linked list:         // 1 -> 2 -> 3 -> 4 -> 5 -> 6         Node head = new Node(1);         head.next = new Node(2);         head.next.next = new Node(3);         head.next.next.next = new Node(4);         head.next.next.next.next = new Node(5);         head.next.next.next.next.next = new Node(6);          head = reverseKGroup(head, 3);          printList(head);     } }
Python 
# Python3 program to reverse a Linked List # in groups of given size  class Node:     def __init__(self, data):         self.data = data         self.next = None  # Function to reverse the linked list in groups def reverseKGroup(head, k):     if not head or k == 1:         return head      st = []     curr = head     prev = None      while curr is not None:                # Terminate the loop when either         # current == None or count >= k         count = 0         while curr is not None and count < k:             st.append(curr)             curr = curr.next             count += 1          # Now pop the elements from the stack one by one         while len(st) > 0:                        # If the final list has not been started yet             if prev is None:                 prev = st.pop()                 head = prev             else:                 prev.next = st.pop()                 prev = prev.next      # Set the next pointer of the last node to None     prev.next = None      return head  def printList(head):     curr = head     while curr is not None:         print(curr.data, end=" ")         curr = curr.next     print()  if __name__ == "__main__":        # Creating a sample singly linked list:     # 1 -> 2 -> 3 -> 4 -> 5 -> 6     head = Node(1)     head.next = Node(2)     head.next.next = Node(3)     head.next.next.next = Node(4)     head.next.next.next.next = Node(5)     head.next.next.next.next.next = Node(6)      head = reverseKGroup(head, 3)      printList(head)
C# 
// C# program to reverse a linked list in  // groups of given size  using System; using System.Collections.Generic;  class Node {     public int data;     public Node next;      public Node(int data) {         this.data = data;         this.next = null;     } }  class GfG {        // Function to reverse the linked list in groups     static Node reverseKGroup(Node head, int k) {         if (head == null || k == 1) {             return head;         }          Stack<Node> st = new Stack<Node>();         Node curr = head;         Node prev = null;          while (curr != null) {                        // Terminate the loop when either           	// current == null or count >= k             int count = 0;             while (curr != null && count < k) {                 st.Push(curr);                 curr = curr.next;                 count++;             }              // Now pop the elements from the            // stack one by one             while (st.Count > 0) {                                // If the final list has not been started yet                 if (prev == null) {                     prev = st.Pop();                     head = prev;                 } else {                     prev.next = st.Pop();                     prev = prev.next;                 }             }         }          // Set the next pointer of the last node to null         prev.next = null;          return head;     }      static void printList(Node head) {         Node curr = head;         while (curr != null) {             Console.Write(curr.data + " ");             curr = curr.next;         }         Console.WriteLine();     }      static void Main(string[] args) {                // Creating a sample singly linked list:         // 1 -> 2 -> 3 -> 4 -> 5 -> 6         Node head = new Node(1);         head.next = new Node(2);         head.next.next = new Node(3);         head.next.next.next = new Node(4);         head.next.next.next.next = new Node(5);         head.next.next.next.next.next = new Node(6);          head = reverseKGroup(head, 3);          printList(head);     } }
JavaScript 
// JavaScript program to reverse a linked list // in groups of given size  class Node {     constructor(data) {         this.data = data;         this.next = null;     } }  // Function to reverse the linked list in groups function reverseKGroup(head, k) {     if (!head || k === 1) {         return head;     }      let st = [];     let curr = head;     let prev = null;      while (curr !== null) {              // Terminate the loop when either          // current == null or count >= k         let count = 0;         while (curr !== null && count < k) {             st.push(curr);             curr = curr.next;             count++;         }          // Now pop the elements from the stack one by one         while (st.length > 0) {                      // If the final list has not been started yet             if (prev === null) {                 prev = st.pop();                 head = prev;             } else {                 prev.next = st.pop();                 prev = prev.next;             }         }     }      // Set the next pointer of the last node to null     prev.next = null;      return head; }  function printList(head) {     let curr = head;     while (curr !== null) {         console.log(curr.data + " ");         curr = curr.next;     }     console.log(); }  // Creating a sample singly linked list: // 1 -> 2 -> 3 -> 4 -> 5 -> 6 let head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); head.next.next.next.next.next = new Node(6);  head = reverseKGroup(head, 3);  printList(head);

Output
3 2 1 6 5 4

Time Complexity: O(n), where n is the number of nodes in the linked list.
Auxiliary Space: O(k)

Related articles:

  • Reverse a Linked List in groups of given size using Deque
  • Reverse a singly Linked List in groups of given size (Recursive Approach)
  • Reverse a singly Linked List in groups of given size (Iterative Approach)


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