Reverse alternate K nodes in a Singly Linked List

3.3

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

#include<stdio.h>
#include<stdlib.h>

/* Link list node */
struct Node
{
    int data;
    struct Node* next;
};

/* Reverses alternate k nodes and
   returns the pointer to the new head node */
struct Node *kAltReverse(struct Node *head, int k)
{
    struct Node* current = head;
    struct Node* next;
    struct 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(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++;
    }
}    

/* Drier 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

// 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


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)

Please write comments if you find the above code/algorithm incorrect, or find other ways to solve the same problem.

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