Reverse a Linked List in groups of given size | Set 1

24 Jun, 2022
Given a linked list, write a function to reverse every k nodes (where k is an input to the function). 
Example: 
Input: 1->2->3->4->5->6->7->8->NULL, K = 3 Output: 3->2->1->6->5->4->8->7->NULL Input: 1->2->3->4->5->6->7->8->NULL, K = 5 Output: 5->4->3->2->1->8->7->6->NULL 
Algorithm: reverse(head, k) 
Reverse the first sub-list of size k. While reversing keep track of the next node and previous node. Let the pointer to the next node be next and pointer to the previous node be prev. See this post for reversing a linked list.
head->next = reverse(next, k) ( Recursively call for rest of the list and link the two sub-lists )
Return prev ( prev becomes the new head of the list (see the diagrams of an iterative method of this post )
Below is image shows how the reverse function works: 
Below is the implementation of the above approach:
C++
C
Java
Python3
C#
Javascript
// CPP program to reverse a linked list// in groups of given size#include <bits/stdc++.h>using namespace std; /* Link list node */class Node {public:    int data;    Node* next;}; /* Reverses the linked list in groupsof size k and returns the pointerto the new head node. */Node* reverse(Node* head, int k){    // base case    if (!head)        return NULL;    Node* current = head;    Node* next = NULL;    Node* prev = NULL;    int count = 0;     /*reverse first k nodes of the linked list */    while (current != NULL && count < k) {        next = current->next;        current->next = prev;        prev = current;        current = next;        count++;    }     /* next is now a pointer to (k+1)th node    Recursively call for the list starting from current.    And make rest of the list as next of first node */    if (next != NULL)        head->next = reverse(next, k);     /* prev is new head of the input list */    return prev;} /* UTILITY FUNCTIONS *//* Function to push a node */void push(Node** head_ref, int new_data){    /* allocate node */    Node* new_node = new Node();     /* put in the data */    new_node->data = new_data;     /* link the old list off the new node */    new_node->next = (*head_ref);     /* move the head to point to the new node */    (*head_ref) = new_node;} /* Function to print linked list */void printList(Node* node){    while (node != NULL) {        cout << node->data << " ";        node = node->next;    }} /* Driver code*/int main(){    /* Start with the empty list */    Node* head = NULL;     /* Created Linked list       is 1->2->3->4->5->6->7->8->9 */    push(&head, 9);    push(&head, 8);    push(&head, 7);    push(&head, 6);    push(&head, 5);    push(&head, 4);    push(&head, 3);    push(&head, 2);    push(&head, 1);     cout << "Given linked list \n";    printList(head);    head = reverse(head, 3);     cout << "\nReversed Linked list \n";    printList(head);     return (0);} // This code is contributed by rathbhupendra
// C program to reverse a linked list in groups of given size#include<stdio.h>#include<stdlib.h> /* Link list node */struct Node{    int data;    struct Node* next;}; /* Reverses the linked list in groups of size k and returns the   pointer to the new head node. */struct Node *reverse (struct Node *head, int k){    if (!head)        return NULL;       struct Node* current = head;    struct Node* next = NULL;    struct Node* prev = NULL;    int count = 0;                 /*reverse first k nodes of the linked list */    while (current != NULL && count < k)    {        next  = current->next;        current->next = prev;        prev = current;        current = next;        count++;    }         /* next is now a pointer to (k+1)th node       Recursively call for the list starting from current.       And make rest of the list as next of first node */    if (next !=  NULL)       head->next = reverse(next, k);     /* prev is new head of the input list */    return prev;} /* UTILITY FUNCTIONS *//* Function to push a node */void push(struct Node** head_ref, int new_data){    /* allocate node */    struct Node* new_node =            (struct Node*) malloc(sizeof(struct Node));     /* put in the data  */    new_node->data  = new_data;     /* link the old list off the new node */    new_node->next = (*head_ref);        /* move the head to point to the new node */    (*head_ref)    = new_node;} /* Function to print linked list */void printList(struct Node *node){    while (node != NULL)    {        printf("%d  ", node->data);        node = node->next;    }}    /* Driver code*/int main(void){    /* Start with the empty list */    struct Node* head = NULL;       /* Created Linked list is 1->2->3->4->5->6->7->8->9 */     push(&head, 9);     push(&head, 8);     push(&head, 7);     push(&head, 6);     push(&head, 5);     push(&head, 4);     push(&head, 3);     push(&head, 2);     push(&head, 1);                printf("\nGiven linked list \n");     printList(head);     head = reverse(head, 3);      printf("\nReversed Linked list \n");     printList(head);      return(0);}
// Java program to reverse a linked list in groups of// given sizeclass LinkedList {    Node head; // head of list     /* Linked list Node*/    class Node {        int data;        Node next;        Node(int d)        {            data = d;            next = null;        }    }     Node reverse(Node head, int k)    {        if(head == null)          return null;        Node current = head;        Node next = null;        Node prev = null;         int count = 0;         /* Reverse first k nodes of linked list */        while (count < k && current != null) {            next = current.next;            current.next = prev;            prev = current;            current = next;            count++;        }         /* next is now a pointer to (k+1)th node           Recursively call for the list starting from           current. And make rest of the list as next of           first node */        if (next != null)            head.next = reverse(next, k);         // prev is now head of input list        return prev;    }     /* Utility functions */     /* Inserts a new Node at front of the list. */    public void push(int new_data)    {        /* 1 & 2: Allocate the Node &                  Put in the data*/        Node new_node = new Node(new_data);         /* 3. Make next of new Node as head */        new_node.next = head;         /* 4. Move the head to point to new Node */        head = new_node;    }     /* Function to print linked list */    void printList()    {        Node temp = head;        while (temp != null) {            System.out.print(temp.data + " ");            temp = temp.next;        }        System.out.println();    }     /* Driver program to test above functions */    public static void main(String args[])    {        LinkedList llist = new LinkedList();         /* Constructed Linked List is 1->2->3->4->5->6->           7->8->8->9->null */        llist.push(9);        llist.push(8);        llist.push(7);        llist.push(6);        llist.push(5);        llist.push(4);        llist.push(3);        llist.push(2);        llist.push(1);         System.out.println("Given Linked List");        llist.printList();         llist.head = llist.reverse(llist.head, 3);         System.out.println("Reversed list");        llist.printList();    }}/* This code is contributed by Rajat Mishra */
# Python program to reverse a# linked list in group of given size # Node class  class Node:     # Constructor to initialize the node object    def __init__(self, data):        self.data = data        self.next = None  class LinkedList:     # Function to initialize head    def __init__(self):        self.head = None     def reverse(self, head, k):               if head == None:          return None        current = head        next = None        prev = None        count = 0         # Reverse first k nodes of the linked list        while(current is not None and count < k):            next = current.next            current.next = prev            prev = current            current = next            count += 1         # next is now a pointer to (k+1)th node        # recursively call for the list starting        # from current. And make rest of the list as        # next of first node        if next is not None:            head.next = self.reverse(next, k)         # prev is new head of the input list        return prev     # Function to insert a new node at the beginning    def push(self, new_data):        new_node = Node(new_data)        new_node.next = self.head        self.head = new_node     # Utility function to print the linked LinkedList    def printList(self):        temp = self.head        while(temp):            print(temp.data,end=' ')            temp = temp.next  # Driver programllist = LinkedList()llist.push(9)llist.push(8)llist.push(7)llist.push(6)llist.push(5)llist.push(4)llist.push(3)llist.push(2)llist.push(1) print("Given linked list")llist.printList()llist.head = llist.reverse(llist.head, 3) print ("\nReversed Linked list")llist.printList() # This code is contributed by Nikhil Kumar Singh(nickzuck_007)
// C# program to reverse a linked list// in groups of given sizeusing System; public class LinkedList {    Node head; // head of list     /* Linked list Node*/    class Node {        public int data;        public Node next;        public Node(int d)        {            data = d;            next = null;        }    }     Node reverse(Node head, int k)    {        if(head == null)          return null;        Node current = head;        Node next = null;        Node prev = null;         int count = 0;         /* Reverse first k nodes of linked list */        while (count < k && current != null) {            next = current.next;            current.next = prev;            prev = current;            current = next;            count++;        }         /* next is now a pointer to (k+1)th node            Recursively call for the list starting from           current. And make rest of the list as next of           first node */        if (next != null)            head.next = reverse(next, k);         // prev is now head of input list        return prev;    }     /* Utility functions */     /* Inserts a new Node at front of the list. */    public void push(int new_data)    {        /* 1 & 2: Allocate the Node &                Put in the data*/        Node new_node = new Node(new_data);         /* 3. Make next of new Node as head */        new_node.next = head;         /* 4. Move the head to point to new Node */        head = new_node;    }     /* Function to print linked list */    void printList()    {        Node temp = head;        while (temp != null) {            Console.Write(temp.data + " ");            temp = temp.next;        }        Console.WriteLine();    }     /* Driver code*/    public static void Main(String[] args)    {        LinkedList llist = new LinkedList();         /* Constructed Linked List is 1->2->3->4->5->6->        7->8->8->9->null */        llist.push(9);        llist.push(8);        llist.push(7);        llist.push(6);        llist.push(5);        llist.push(4);        llist.push(3);        llist.push(2);        llist.push(1);         Console.WriteLine("Given Linked List");        llist.printList();         llist.head = llist.reverse(llist.head, 3);         Console.WriteLine("Reversed list");        llist.printList();    }} // This code is contributed by 29AjayKumar
<script> // Javascript program to reverse a// linked list in groups of// given sizevar head; // head of list     /* Linked list Node */     class Node {            constructor(val) {                this.data = val;                this.next = null;            }        }     function reverse(head , k) {        if (head == null)            return null;        var current = head;        var next = null;        var prev = null;         var count = 0;         /* Reverse first k nodes of linked list */        while (count < k && current != null) {            next = current.next;            current.next = prev;            prev = current;            current = next;            count++;        }         /*         next is now a pointer to (k+1)th node         Recursively call for the list starting         from current. And make rest of the list         as next of first node         */        if (next != null)            head.next = reverse(next, k);         // prev is now head of input list        return prev;    }     /* Utility functions */     /* Inserts a new Node at front of the list. */    function push(new_data) {        /*         1 & 2: Allocate the Node & Put in the data         */        new_node = new Node(new_data);         /* 3. Make next of new Node as head */        new_node.next = head;         /* 4. Move the head to point to new Node */        head = new_node;    }     /* Function to print linked list */    function printList() {        temp = head;        while (temp != null) {            document.write(temp.data + " ");            temp = temp.next;        }        document.write("<br/>");    }     /* Driver program to test above functions */                      /*          Constructed Linked List is          1->2->3->4->5->6-> 7->8->8->9->null         */        push(9);        push(8);        push(7);        push(6);        push(5);        push(4);        push(3);        push(2);        push(1);         document.write("Given Linked List<br/>");        printList();         head = reverse(head, 3);         document.write("Reversed list<br/>");        printList(); // This code contributed by gauravrajput1 </script>
Given linked list 
1 2 3 4 5 6 7 8 9 
Reversed Linked list 
3 2 1 6 5 4 9 8 7 
Complexity Analysis: 
Time Complexity: O(n). Traversal of list is done only once and it has ‘n’ elements.
Auxiliary Space: O(n/k). For each Linked List of size n, n/k or (n/k)+1 calls will be made during the recursion.
We can solve this question in O(1) Space Complexity.
The following steps are required for this Algorithm:
Create a dummy node and point it to the head of input i.e dummy->next = head. Calculate the length of the linked list which takes O(N) time, where N is the length of the linked list. Initialize three-pointers prev, curr, next to reverse k elements for every group.Iterate over the linked lists till next!=NULL.Points curr to the prev->next and next to the curr next.Then, Using the inner for loop reverse the particular group using these four steps:
Create a dummy node and point it to the head of input i.e dummy->next = head.
 Calculate the length of the linked list which takes O(N) time, where N is the length of the linked list.
 Initialize three-pointers prev, curr, next to reverse k elements for every group.
Iterate over the linked lists till next!=NULL.
Points curr to the prev->next and next to the curr next.
Then, Using the inner for loop reverse the particular group using these four steps:
curr->next = next->next
next->next = prev->next
prev->next = next
next = curr->next
       7. This for loop runs for k-1 times for all groups except the last remaining element, for the last remaining element it runs              for the remaining length of the linked list – 1.
       8. Decrement count after for loop by k count -= k, to determine the length of the remaining linked list.
       9. Change prev position to curr, prev = curr.
Here is the code for the above algorithm.
C++
C
Java
Python3
C#
// CPP program to reverse a linked list// in groups of given size#include <bits/stdc++.h>using namespace std; /* Link list node */class Node {public:    int data;    Node* next;}; /* Reverses the linked list in groupsof size k and returns the pointerto the new head node. */Node* reverse(Node* head, int k){    // If head is NULL or K is 1 then return head    if (!head || k == 1)        return head;     Node* dummy = new Node(); // creating dummy node    dummy->data = -1;    dummy->next = head;     // Initializing three points prev, curr, next    Node *prev = dummy, *curr = dummy, *next = dummy;     // Calculating the length of linked list    int count = 0;    while (curr) {        curr = curr->next;        count++;    }     // Iterating till next is not NULL    while (next) {        // Curr position after every reverse group        curr = prev->next;        // Next will always next to curr        next = curr->next;        // toLoop will set to count - 1 in case of remaining        // element        int toLoop = count > k ? k : count - 1;        for (int i = 1; i < toLoop; i++) {            // 4 steps as discussed above            curr->next = next->next;            next->next = prev->next;            prev->next = next;            next = curr->next;        }        // Setting prev to curr        prev = curr;        // Update count        count -= k;    }    // dummy -> next will be our new head for output linked    // list    return dummy->next;} /* UTILITY FUNCTIONS *//* Function to push a node */void push(Node** head_ref, int new_data){    /* allocate node */    Node* new_node = new Node();     /* put in the data */    new_node->data = new_data;     /* link the old list off the new node */    new_node->next = (*head_ref);     /* move the head to point to the new node */    (*head_ref) = new_node;} /* Function to print linked list */void printList(Node* node){    while (node != NULL) {        cout << node->data << " ";        node = node->next;    }} /* Driver code*/int main(){    /* Start with the empty list */    Node* head = NULL;     /* Created Linked list    is 1->2->3->4->5->6->7->8->9 */    push(&head, 9);    push(&head, 8);    push(&head, 7);    push(&head, 6);    push(&head, 5);    push(&head, 4);    push(&head, 3);    push(&head, 2);    push(&head, 1);     cout << "Given linked list \n";    printList(head);    head = reverse(head, 3);     cout << "\nReversed Linked list \n";    printList(head);     return (0);} // This code is contributed by Sania Kumari Gupta// (kriSania804)
// C program to reverse a linked list// in groups of given size#include <stdio.h>#include <stdlib.h> /* Link list node */typedef struct Node {    int data;    struct Node* next;} Node; // Reverses the linked list in groups of size k and returns// the pointer to the new head node.Node* reverse(Node* head, int k){    // If head is NULL or K is 1 then return head    if (!head || k == 1)        return head;    // creating dummy node    Node* dummy = (Node*)malloc(sizeof(Node));    dummy->data = -1;    dummy->next = head;     // Initializing three points prev, curr, next    Node *prev = dummy, *curr = dummy, *next = dummy;     // Calculating the length of linked list    int count = 0;    while (curr) {        curr = curr->next;        count++;    }     // Iterating till next is not NULL    while (next) {        // Curr position after every reverse group        curr = prev->next;        // Next will always next to curr        next = curr->next;        // toLoop will set to count - 1 in case of remaining        // element        int toLoop = count > k ? k : count - 1;        for (int i = 1; i < toLoop; i++) {            // 4 steps as discussed above            curr->next = next->next;            next->next = prev->next;            prev->next = next;            next = curr->next;        }        // Setting prev to curr        prev = curr;        // Update count        count -= k;    }    // dummy -> next will be our new head for output linked    // list    return dummy->next;} /* UTILITY FUNCTIONS *//* Function to push a node */void push(Node** head_ref, int new_data){    /* allocate node */    Node* new_node = (Node*)malloc(sizeof(Node));     /* put in the data */    new_node->data = new_data;     /* link the old list off the new node */    new_node->next = (*head_ref);     /* move the head to point to the new node */    (*head_ref) = new_node;} /* Function to print linked list */void printList(Node* node){    while (node != NULL) {        printf("%d ", node->data);        node = node->next;    }} /* Driver code*/int main(){    /* Start with the empty list */    Node* head = NULL;     /* Created Linked list    is 1->2->3->4->5->6->7->8->9 */    push(&head, 9);    push(&head, 8);    push(&head, 7);    push(&head, 6);    push(&head, 5);    push(&head, 4);    push(&head, 3);    push(&head, 2);    push(&head, 1);     printf("Given linked list \n");    printList(head);    head = reverse(head, 3);     printf("\nReversed Linked list \n");    printList(head);     return (0);} // This code is contributed by Sania Kumari Gupta// (kriSania804)
import java.util.*; // Linked List Nodeclass Node {    int data;    Node next;    Node(int a)    {        data = a;        next = null;    }} class GFG {    // utility function to insert node in the list    static Node push(Node head, int val)    {        Node newNode = new Node(val);        if (head == null) {            head = newNode;            return head;        }         Node temp = head;        while (temp.next != null)            temp = temp.next;         temp.next = newNode;        return head;    }     // utility function to reverse k nodes in the list    static Node reverse(Node head, int k)    {        // If head is NULL or K is 1 then return head        if (head == null || head.next == null)            return head;         // creating dummy node        Node dummy = new Node(-1);        dummy.next = head;         // Initializing three points prev, curr, next        Node prev = dummy;        Node curr = dummy;        Node next = dummy;         // Calculating the length of linked list        int count = 0;        while (curr != null) {            count++;            curr = curr.next;        }         // Iterating till next is not NULL        while (next != null) {            curr = prev.next; // Curr position after every                              // reverse group            next = curr.next; // Next will always next to                              // curr            int toLoop                = count > k                      ? k                      : count - 1; // toLoop will set to                                   // count - 1 in case of                                   // remaining element             for (int i = 1; i < toLoop; i++) {                // 4 steps as discussed above                curr.next = next.next;                next.next = prev.next;                prev.next = next;                next = curr.next;            }            prev = curr; // Setting prev to curr            count -= k; // Update count        }        return dummy.next; // dummy -> next will be our new                           // head for output linked        // list    }    // utility function to print the list    static void print(Node head)    {        while (head.next != null) {            System.out.print(head.data + " ");            head = head.next;        }        System.out.println(head.data);    }     public static void main(String args[])    {        Node head = null;        int k = 3;        head = push(head, 1);        head = push(head, 2);        head = push(head, 3);        head = push(head, 4);        head = push(head, 5);        head = push(head, 6);        head = push(head, 7);        head = push(head, 8);        head = push(head, 9);         System.out.println("Given Linked List");        print(head);        System.out.println("Reversed list");        Node newHead = reverse(head, k);        print(newHead);    }}
# Python program to reverse a linked list in groups of given sizeclass Node:    def __init__(self, data):        self.data = data        self.next = None # Reverses the linked list in groups# of size k and returns the pointer# to the new head node.  def reverse(head, k):  # If head is NULL or K is 1 then return head    if not head or k == 1:        return head    dummy = Node(-1)  # creating dummy node    dummy.next = head    # Initializing three points prev, curr, next    prev = dummy    curr = dummy    next = dummy    count = 0    toLoop = 0    i = 0     # Calculating the length of linked list    while curr:        curr = curr.next        count += 1     # Iterating till next is not none    while next:        curr = prev.next  # Curr position after every reversed group        next = curr.next  # Next will always next to curr        # toLoop will set to count - 1 in case of remaining element        toLoop = count > k and k or count - 1        for i in range(1, toLoop):                # 4 steps as discussed above            curr.next = next.next            next.next = prev.next            prev.next = next            next = curr.next        # Setting prev to curr        prev = curr        # Update count        count -= k      # dummy -> next will be our new head for output linked list    return dummy.next # Function to print linked list  def printList(node):    while node is not None:        print(node.data, end=" ")        node = node.next  # Created Linked list is 1->2->3->4->5->6->7->8->9head = 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.next.next.next.next.next.next = Node(7)head.next.next.next.next.next.next.next = Node(8)head.next.next.next.next.next.next.next.next = Node(9) print("Given linked list")printList(head)head = reverse(head, 3) print("\nReversed Linked list")printList(head) # This code is contributed by Tapesh(tapeshdua420)
// C# program to reverse a linked list// in groups of given size using System; /* Link list node */class Node {    public int data;    public Node next;    public Node(int a)    {        data = a;        next = null;    }} class GFG {    /* UTILITY FUNCTIONS */    /* Function to push a node */    static Node push(Node head, int val)    {         Node newNode = new Node(val);        if (head == null) {            head = newNode;            return head;        }         Node temp = head;        while (temp.next != null)            temp = temp.next;         temp.next = newNode;        return head;    }     // utility function to reverse k nodes in the list    static Node reverse(Node head, int k)    {        // If head is NULL or K is 1 then return head        if (head == null || head.next == null)            return head;         // creating dummy node        Node dummy = new Node(-1);        dummy.next = head;         // Initializing three points prev, curr, next        Node prev = dummy;        Node curr = dummy;        Node next = dummy;         // Calculating the length of linked list        int count = 0;        while (curr != null) {            count++;            curr = curr.next;        }         // Iterating till next is not NULL        while (next != null) {            curr = prev.next; // Curr position after every                              // reverse group            next = curr.next; // Next will always next to                              // curr             int toLoop                = count > k                      ? k                      : count - 1; // toLoop will set to                                   // count - 1 in case of                                   // remaining element             for (int i = 1; i < toLoop; i++) {                // 4 steps as discussed above                curr.next = next.next;                next.next = prev.next;                prev.next = next;                next = curr.next;            }             prev = curr; // Setting prev to curr            count -= k; // Update count        }         return dummy.next; // dummy -> next will be our new                           // head for output linked list    }     // utility function to print the list    static void print(Node head)    {        while (head.next != null) {            Console.Write(head.data + " ");            head = head.next;        }        Console.WriteLine(head.data);    }    public static void Main()    {        Node head = null;        int K = 3;        head = push(head, 1);        head = push(head, 2);        head = push(head, 3);        head = push(head, 4);        head = push(head, 5);        head = push(head, 6);        head = push(head, 7);        head = push(head, 8);        head = push(head, 9);        Console.WriteLine("Given linked list");        print(head);        Console.WriteLine("Reversed Linked list");        Node newHead = reverse(head, K);        print(newHead);    }} // This code is contributed by Tapesh (tapeshdua420)
Given linked list 
1 2 3 4 5 6 7 8 9 
Reversed Linked list 
3 2 1 6 5 4 9 8 7 
Time Complexity: O(N) : While loop takes O(N/K) time and inner for loop takes O(K) time. So N/K * K = N. Therefore TC O(N)
Space Complexity: O(1) : No extra space is used.
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Please write comments if you find the above code/algorithm incorrect, or find other ways to solve the same problem. 
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