Reverse alternate K nodes in a Singly Linked List

03 Jul, 2022
Given a linked list, write a function to reverse every alternate k nodes (where k is an input to the function) in an efficient way. Give the complexity of your algorithm.
Example: 
Inputs:   1->2->3->4->5->6->7->8->9->NULL and k = 3
Output:   3->2->1->4->5->6->9->8->7->NULL. 
Method 1 (Process 2k nodes and recursively call for rest of the list) 
This method is basically an extension of the method discussed in this post. 
kAltReverse(struct node *head, int k)
  1)  Reverse first k nodes.
  2)  In the modified list head points to the kth node.  So change next 
       of head to (k+1)th node
  3)  Move the current pointer to skip next k nodes.
  4)  Call the kAltReverse() recursively for rest of the n - 2k nodes.
  5)  Return new head of the list.
C++
Java
Python3
C#
Javascript
// C++ program to reverse alternate// k nodes in a linked list#include <bits/stdc++.h>using namespace std;  /* Link list node */class Node {     public:    int data;     Node* next; };   /* Reverses alternate k nodes and returns the pointer to the new head node */Node *kAltReverse(Node *head, int k) {     Node* current = head;     Node* next;     Node* prev = NULL;     int count = 0;       /*1) reverse first k nodes of the linked list */    while (current != NULL && count < k)     {     next = current->next;     current->next = prev;     prev = current;     current = next;     count++;     }           /* 2) Now head points to the kth node.     So change next  of head to (k+1)th node*/    if(head != NULL)     head->next = current;       /* 3) We do not want to reverse next k        nodes. So move the current         pointer to skip next k nodes */    count = 0;     while(count < k-1 && current != NULL )     {     current = current->next;     count++;     }       /* 4) Recursively call for the list     starting from current->next. And make    rest of the list as next of first node */    if(current != NULL)     current->next = kAltReverse(current->next, k);       /* 5) 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) {     int count = 0;     while(node != NULL)     {         cout<<node->data<<" ";         node = node->next;         count++;     } }   /* Driver code*/int main(void) {     /* Start with the empty list */    Node* head = NULL;     int i;           // create a list 1->2->3->4->5...... ->20     for(i = 20; i > 0; i--)     push(&head, i);       cout<<"Given linked list \n";     printList(head);     head = kAltReverse(head, 3);       cout<<"\n Modified Linked list \n";     printList(head);     return(0); }     // This code is contributed by rathbhupendra
// Java program to reverse alternate k nodes in a linked list  class LinkedList {      static Node head;      class Node {          int data;        Node next;          Node(int d) {            data = d;            next = null;        }    }      /* Reverses alternate k nodes and     returns the pointer to the new head node */    Node kAltReverse(Node node, int k) {        Node current = node;        Node next = null, prev = null;        int count = 0;          /*1) reverse first k nodes of the linked list */        while (current != null && count < k) {            next = current.next;            current.next = prev;            prev = current;            current = next;            count++;        }          /* 2) Now head points to the kth node.  So change next          of head to (k+1)th node*/        if (node != null) {            node.next = current;        }          /* 3) We do not want to reverse next k nodes. So move the current          pointer to skip next k nodes */        count = 0;        while (count < k - 1 && current != null) {            current = current.next;            count++;        }          /* 4) Recursively call for the list starting from current->next.         And make rest of the list as next of first node */        if (current != null) {            current.next = kAltReverse(current.next, k);        }          /* 5) prev is new head of the input list */        return prev;    }      void printList(Node node) {        while (node != null) {            System.out.print(node.data + " ");            node = node.next;        }    }      void push(int newdata) {        Node mynode = new Node(newdata);        mynode.next = head;        head = mynode;    }      public static void main(String[] args) {        LinkedList list = new LinkedList();          // Creating the linkedlist        for (int i = 20; i > 0; i--) {            list.push(i);        }        System.out.println("Given Linked List :");        list.printList(head);        head = list.kAltReverse(head, 3);        System.out.println("");        System.out.println("Modified Linked List :");        list.printList(head);      }}  // This code has been contributed by Mayank Jaiswal
# Python3 program to reverse alternate# k nodes in a linked listimport math  # Link list node class Node:     def __init__(self, data):         self.data = data         self.next = None  # Reverses alternate k nodes and #returns the pointer to the new head node def kAltReverse(head, k) :     current = head     next = None    prev = None    count = 0      #1) reverse first k nodes of the linked list     while (current != None and count < k) :         next = current.next        current.next = prev         prev = current         current = next        count = count + 1;          # 2) Now head pos to the kth node.     # So change next of head to (k+1)th node    if(head != None):         head.next = current       # 3) We do not want to reverse next k     # nodes. So move the current     # pointer to skip next k nodes     count = 0    while(count < k - 1 and current != None ):         current = current.next        count = count + 1          # 4) Recursively call for the list     # starting from current.next. And make    # rest of the list as next of first node     if(current != None):         current.next = kAltReverse(current.next, k)       # 5) prev is new head of the input list     return prev   # UTILITY FUNCTIONS # Function to push a node def push(head_ref, new_data):           # allocate node     new_node = Node(new_data)      # 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 po to the new node     head_ref = new_node          return head_ref  # Function to print linked list def printList(node):     count = 0    while(node != None):         print(node.data, end = " ")         node = node.next        count = count + 1      # Driver codeif __name__=='__main__':          # Start with the empty list     head = None      # create a list 1.2.3.4.5...... .20     for i in range(20, 0, -1):        head = push(head, i)               print("Given linked list ")     printList(head)     head = kAltReverse(head, 3)       print("\nModified Linked list")     printList(head)       # This code is contributed by Srathore
// C# program to reverse alternate// k nodes in a linked list using System;class LinkedList {       static Node head;       public class Node     {           public int data;         public Node next;           public Node(int d)        {             data = d;             next = null;         }     }       /* Reverses alternate k nodes and     returns the pointer to the new head node */    Node kAltReverse(Node node, int k)     {         Node current = node;         Node next = null, prev = null;         int count = 0;           /*1) reverse first k nodes of the linked list */        while (current != null && count < k)        {             next = current.next;             current.next = prev;             prev = current;             current = next;             count++;         }           /* 2) Now head points to the kth         node. So change next         of head to (k+1)th node*/        if (node != null)         {             node.next = current;         }           /* 3) We do not want to reverse         next k nodes. So move the current         pointer to skip next k nodes */        count = 0;         while (count < k - 1 && current != null)         {             current = current.next;             count++;         }           /* 4) Recursively call for the         list starting from current->next.         And make rest of the list as         next of first node */        if (current != null)        {             current.next = kAltReverse(current.next, k);         }           /* 5) prev is new head of the input list */        return prev;     }       void printList(Node node)     {         while (node != null)         {             Console.Write(node.data + " ");             node = node.next;         }     }       void push(int newdata)    {         Node mynode = new Node(newdata);         mynode.next = head;         head = mynode;     }       // Driver code    public static void Main(String []args)    {         LinkedList list = new LinkedList();           // Creating the linkedlist         for (int i = 20; i > 0; i--)         {             list.push(i);         }         Console.WriteLine("Given Linked List :");         list.printList(head);         head = list.kAltReverse(head, 3);         Console.WriteLine("");         Console.WriteLine("Modified Linked List :");         list.printList(head);     } }   // This code has been contributed by Arnab Kundu
<script>  // JavaScript program to reverse // alternate k nodes in a linked listclass Node {    constructor(d)    {        this.data = d;        this.next = null;    }}  let head;  // Reverses alternate k nodes and returns// the pointer to the new head node function kAltReverse(node, k){    let current = node;    let next = null, prev = null;    let count = 0;      /*1) reverse first k nodes of the linked list */    while (current != null && count < k)    {        next = current.next;        current.next = prev;        prev = current;        current = next;        count++;    }      /* 2) Now head points to the kth node.            So change next of head to (k+1)th node*/    if (node != null)    {        node.next = current;    }      /* 3) We do not want to reverse next k nodes.          So move the current pointer to skip           next k nodes */    count = 0;    while (count < k - 1 && current != null)     {        current = current.next;        count++;    }      /* 4) Recursively call for the list starting          from current->next. And make rest of           the list as next of first node */    if (current != null)     {        current.next = kAltReverse(current.next, k);    }      /* 5) prev is new head of the input list */    return prev;}  function printList(node){    while (node != null)    {        document.write(node.data + " ");        node = node.next;    }}  function push(newdata){    let mynode = new Node(newdata);    mynode.next = head;    head = mynode;}  // Driver code  // Creating the linkedlistfor(let i = 20; i > 0; i--){    push(i);}document.write("Given Linked List :<br>");printList(head);head = kAltReverse(head, 3);  document.write("<br>");document.write("Modified Linked List :<br>");printList(head);  // This code is contributed by rag2127  </script>
Output: 
Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19
Time Complexity: O(n)
Space Complexity: O(n)
Method 2 (Process k nodes and recursively call for rest of the list) 
The method 1 reverses the first k node and then moves the pointer to k nodes ahead. So method 1 uses two while loops and processes 2k nodes in one recursive call. 
This method processes only k nodes in a recursive call. It uses a third bool parameter b which decides whether to reverse the k elements or simply move the pointer.  
_kAltReverse(struct node *head, int k, bool b)
  1)  If b is true, then reverse first k nodes.
  2)  If b is false, then move the pointer k nodes ahead.
  3)  Call the kAltReverse() recursively for rest of the n - k nodes and link 
       rest of the modified list with end of first k nodes. 
  4)  Return new head of the list.
C++
C
Java
Python3
C#
Javascript
#include <bits/stdc++.h>using namespace std;  /* Link list node */class node {     public:    int data;     node* next; };   /* Helper function for kAltReverse() */node * _kAltReverse(node *node, int k, bool b);   /* Alternatively reverses the given linked list in groups of given size k. */node *kAltReverse(node *head, int k) {     return _kAltReverse(head, k, true); }   /* Helper function for kAltReverse().  It reverses k nodes of the list only if the third parameter b is passed as true, otherwise moves the pointer k nodes aheadand recursively calls itself */node * _kAltReverse(node *Node, int k, bool b) {     if(Node == NULL)         return NULL;           int count = 1;     node *prev = NULL;     node *current = Node;     node *next;           /* The loop serves two purposes         1) If b is true,            then it reverses the k nodes         2) If b is false,            then it moves the current pointer */    while(current != NULL && count <= k)     {         next = current->next;               /* Reverse the nodes only if b is true*/        if(b == true)             current->next = prev;                           prev = current;         current = next;         count++;     }               /* 3) If b is true, then node is the kth node.         So attach rest of the list after node.         4) After attaching, return the new head */    if(b == true)     {         Node->next = _kAltReverse(current, k, !b);         return prev;             }               /* If b is not true, then attach     rest of the list after prev.     So attach rest of the list after prev */    else    {         prev->next = _kAltReverse(current, k, !b);         return Node;         } }     /* 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) {     int count = 0;     while(node != NULL)     {         cout << node->data << " ";         node = node->next;         count++;     } }   // Driver Codeint main(void) {     /* Start with the empty list */    node* head = NULL;     int i;       // create a list 1->2->3->4->5...... ->20     for(i = 20; i > 0; i--)     push(&head, i);       cout << "Given linked list \n";     printList(head);     head = kAltReverse(head, 3);       cout << "\nModified Linked list \n";     printList(head);     return(0); }   // This code is contributed by rathbhupendra
#include<stdio.h>#include<stdlib.h>  /* Link list node */struct node{    int data;    struct node* next;};  /* Helper function for kAltReverse() */struct node * _kAltReverse(struct node *node, int k, bool b);  /* Alternatively reverses the given linked list in groups of    given size k. */struct node *kAltReverse(struct node *head, int k){  return _kAltReverse(head, k, true);}   /*  Helper function for kAltReverse().  It reverses k nodes of the list only if    the third parameter b is passed as true, otherwise moves the pointer k     nodes ahead and recursively calls itself  */ struct node * _kAltReverse(struct node *node, int k, bool b){   if(node == NULL)       return NULL;     int count = 1;   struct node *prev = NULL;   struct node  *current = node;   struct node *next;      /* The loop serves two purposes      1) If b is true, then it reverses the k nodes       2) If b is false, then it moves the current pointer */   while(current != NULL && count <= k)   {       next = current->next;         /* Reverse the nodes only if b is true*/       if(b == true)          current->next = prev;                     prev = current;       current = next;       count++;   }        /* 3) If b is true, then node is the kth node.        So attach rest of the list after node.      4) After attaching, return the new head */   if(b == true)   {        node->next = _kAltReverse(current,k,!b);        return prev;           }        /* If b is not true, then attach rest of the list after prev.      So attach rest of the list after prev */       else   {        prev->next = _kAltReverse(current, k, !b);        return node;          }}     /* 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){    int count = 0;    while(node != NULL)    {        printf("%d  ", node->data);        node = node->next;        count++;    }}   /* Driver program to test above function*/int main(void){    /* Start with the empty list */    struct node* head = NULL;    int i;       // create a list 1->2->3->4->5...... ->20    for(i = 20; i > 0; i--)      push(&head, i);       printf("\n Given linked list \n");    printList(head);    head = kAltReverse(head, 3);       printf("\n Modified Linked list \n");    printList(head);       getchar();    return(0);}
// Java program to reverse alternate k nodes in a linked list  class LinkedList {      static Node head;      class Node {          int data;        Node next;          Node(int d) {            data = d;            next = null;        }    }      /* Alternatively reverses the given linked list in groups of      given size k. */    Node kAltReverse(Node head, int k) {        return _kAltReverse(head, k, true);    }      /*  Helper function for kAltReverse().  It reverses k nodes of the list only if     the third parameter b is passed as true, otherwise moves the pointer k      nodes ahead and recursively calls itself  */    Node _kAltReverse(Node node, int k, boolean b) {        if (node == null) {            return null;        }          int count = 1;        Node prev = null;        Node current = node;        Node next = null;          /* The loop serves two purposes         1) If b is true, then it reverses the k nodes          2) If b is false, then it moves the current pointer */        while (current != null && count <= k) {            next = current.next;              /* Reverse the nodes only if b is true*/            if (b == true) {                current.next = prev;            }              prev = current;            current = next;            count++;        }          /* 3) If b is true, then node is the kth node.          So attach rest of the list after node.          4) After attaching, return the new head */        if (b == true) {            node.next = _kAltReverse(current, k, !b);            return prev;        } /* If b is not true, then attach rest of the list after prev.          So attach rest of the list after prev */ else {            prev.next = _kAltReverse(current, k, !b);            return node;        }    }      void printList(Node node) {        while (node != null) {            System.out.print(node.data + " ");            node = node.next;        }    }      void push(int newdata) {        Node mynode = new Node(newdata);        mynode.next = head;        head = mynode;    }      public static void main(String[] args) {        LinkedList list = new LinkedList();          // Creating the linkedlist        for (int i = 20; i > 0; i--) {            list.push(i);        }        System.out.println("Given Linked List :");        list.printList(head);        head = list.kAltReverse(head, 3);        System.out.println("");        System.out.println("Modified Linked List :");        list.printList(head);      }}  // This code has been contributed by Mayank Jaiswal
# Python code for above algorithm  # Link list node class node:           def __init__(self, data):         self.data = data         self.next = next          # function to insert a node at the# beginning of the linked listdef push(head_ref, new_data):      # allocate node     new_node = node(0)      # put in the data     new_node.data = new_data      # link the old list to the new node     new_node.next = (head_ref)      # move the head to point to the new node     (head_ref) = new_node          return head_ref      """ Alternatively reverses the given linked list in groups of given size k. """def kAltReverse(head, k) :      return _kAltReverse(head, k, True)   """ Helper function for kAltReverse(). It reverses k nodes of the list only if the third parameter b is passed as True, otherwise moves the pointer k nodes aheadand recursively calls itself """def _kAltReverse(Node, k, b) :      if(Node == None) :        return None          count = 1    prev = None    current = Node     next = None          """ The loop serves two purposes         1) If b is True,         then it reverses the k nodes         2) If b is false,         then it moves the current pointer """    while(current != None and count <= k) :              next = current.next              """ Reverse the nodes only if b is True"""        if(b == True) :            current.next = prev                           prev = current         current = next        count = count + 1                    """ 3) If b is True, then node is the kth node.         So attach rest of the list after node.         4) After attaching, return the new head """    if(b == True) :              Node.next = _kAltReverse(current, k, not b)         return prev                   else :        """ If b is not True, then attach         rest of the list after prev.         So attach rest of the list after prev """        prev.next = _kAltReverse(current, k, not b)         return Node           """ Function to print linked list """def printList(node) :      count = 0    while(node != None) :              print( node.data, end = " ")         node = node.next        count = count + 1  # Driver Code  """ Start with the empty list """head = Nonei = 20  # create a list 1->2->3->4->5...... ->20 while(i > 0 ):     head = push(head, i)     i = i - 1  print( "Given linked list ") printList(head) head = kAltReverse(head, 3)   print( "\nModified Linked list ") printList(head)   # This code is contributed by Arnab Kundu
// C# Program for converting // singly linked list into // circular linked list. using System;  public class LinkedList{       static Node head;       public class Node     {           public int data;         public Node next;           public Node(int d)         {             data = d;             next = null;         }     }       /* Reverses alternate k nodes and     returns the pointer to the new head node */    Node kAltReverse(Node node, int k)     {         Node current = node;         Node next = null, prev = null;         int count = 0;           /*1) reverse first k nodes of the linked list */        while (current != null && count < k)         {             next = current.next;             current.next = prev;             prev = current;             current = next;             count++;         }           /* 2) Now head points to the kth node.         So change next of head to (k+1)th node*/        if (node != null)        {             node.next = current;         }           /* 3) We do not want to reverse next         k nodes. So move the current         pointer to skip next k nodes */        count = 0;         while (count < k - 1 && current != null)        {             current = current.next;             count++;         }           /* 4) Recursively call for the list         starting from current->next. And make         rest of the list as next of first node */        if (current != null)         {             current.next = kAltReverse(current.next, k);         }           /* 5) prev is new head of the input list */        return prev;     }       void printList(Node node)     {         while (node != null)         {             Console.Write(node.data + " ");             node = node.next;         }     }       void push(int newdata)    {         Node mynode = new Node(newdata);         mynode.next = head;         head = mynode;     }       public static void Main(String[] args)     {         LinkedList list = new LinkedList();           // Creating the linkedlist         for (int i = 20; i > 0; i--)        {             list.push(i);         }         Console.WriteLine("Given Linked List :");         list.printList(head);         head = list.kAltReverse(head, 3);         Console.WriteLine("");         Console.WriteLine("Modified Linked List :");         list.printList(head);     } }   // This code is contributed 29AjayKumar
<script>// javascript program to reverse alternate k nodes in a linked listvar head;       class Node {            constructor(val) {                this.data = val;                this.next = null;            }        }             /*     * Alternatively reverses the given linked list in groups of given size k.     */    function kAltReverse(head , k) {        return _kAltReverse(head, k, true);    }      /*     * Helper function for kAltReverse(). It reverses k nodes of the list only if     * the third parameter b is passed as true, otherwise moves the pointer k nodes     * ahead and recursively calls itself     */    function _kAltReverse(node , k,  b) {        if (node == null) {            return null;        }          var count = 1;var prev = null;var current = node;var next = null;          /*         * The loop serves two purposes 1) If b is true, then it reverses the k nodes 2)         * If b is false, then it moves the current pointer         */        while (current != null && count <= k) {            next = current.next;              /* Reverse the nodes only if b is true */            if (b == true) {                current.next = prev;            }              prev = current;            current = next;            count++;        }          /*         * 3) If b is true, then node is the kth node. So attach rest of the list after         * node. 4) After attaching, return the new head         */        if (b == true) {            node.next = _kAltReverse(current, k, !b);            return prev;        } /*             * If b is not true, then attach rest of the list after prev. So attach rest of             * the list after prev             */ else {            prev.next = _kAltReverse(current, k, !b);            return node;        }    }      function printList(node) {        while (node != null) {            document.write(node.data + " ");            node = node.next;        }    }      function push(newdata) {var mynode = new Node(newdata);        mynode.next = head;        head = mynode;    }                        // Creating the linkedlist        for (i = 20; i > 0; i--) {            push(i);        }        document.write("Given Linked List :<br/>");        printList(head);        head = kAltReverse(head, 3);        document.write("<br/>");        document.write("Modified Linked List :<br/>");        printList(head);    // This code contributed by aashish1995</script>
Output: 
Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19
Time Complexity: O(n) 
Space Complexity: O(n)
Please write comments if you find the above code/algorithm incorrect, or find other ways to solve the same problem.
29AjayKumar
andrew1234
rathbhupendra
sapnasingh4991
Akanksha_Rai
nidhi_biet
aashish1995
rag2127
simmytarika5
ashutoshsinghgeeksforgeeks
germanshephered48
surinderdawra388
hardikkoriintern
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