Rearrange a Linked List in Zig-Zag fashion

Given a linked list, rearrange it such that converted list should be of the form a < b > c < d > e < f .. where a, b, c.. are consecutive data node of linked list.

Examples :

Input:  1->2->3->4
Output: 1->3->2->4 

Input:  11->15->20->5->10
Output: 11->20->5->15->10



We strongly recommend that you click here and practice it, before moving on to the solution.

A simple approach to do this, is to sort the linked list using merge sort and then swap alternate, but that requires O(n Log n) time complexity. Here n is number of elements in linked list.

An efficient approach which requires O(n) time is, using a single scan similar to bubble sort and then maintain a flag for representing which order () currently we are. If the current two elements are not in that order then swap those elements otherwise not. Please refer this for detailed explanation of swapping order.

C++

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// C++ program to arrange linked list in zigzag fashion
#include <bits/stdc++.h>
using namespace std;
  
/* Link list Node */
struct Node
{
    int data;
    struct Node* next;
};
  
// This function distributes the Node in zigzag fashion
void zigZagList(Node *head)
{
    // If flag is true, then next node should be greater
    // in the desired output.
    bool flag = true;
  
    // Traverse linked list starting from head.
    Node* current = head;
    while (current->next != NULL)
    {
        if (flag)  /* "<" relation expected */
        {
            /* If we have a situation like A > B > C
               where A, B and C are consecutive Nodes
               in list we get A > B < C by swapping B
               and C */
            if (current->data > current->next->data)
                swap(current->data, current->next->data);
        }
        else /* ">" relation expected */
        {
            /* If we have a situation like A < B < C where
               A, B and C  are consecutive Nodes in list we
               get A < C > B by swapping B and C */
            if (current->data < current->next->data)
                swap(current->data, current->next->data);
        }
  
        current = current->next;
        flag = !flag;  /* flip flag for reverse checking */
    }
}
  
/* UTILITY FUNCTIONS */
/* Function to push a Node */
void push(Node** head_ref, int new_data)
{
    /* allocate Node */
    struct 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(struct Node *Node)
{
    while (Node != NULL)
    {
        printf("%d->", Node->data);
        Node = Node->next;
    }
    printf("NULL");
}
  
/* Driver program to test above function*/
int main(void)
{
    /* Start with the empty list */
    struct Node* head = NULL;
  
    // create a list 4 -> 3 -> 7 -> 8 -> 6 -> 2 -> 1
    // answer should be -> 3  7  4  8  2  6  1
    push(&head, 1);
    push(&head, 2);
    push(&head, 6);
    push(&head, 8);
    push(&head, 7);
    push(&head, 3);
    push(&head, 4);
  
    printf("Given linked list \n");
    printList(head);
  
    zigZagList(head);
  
    printf("\nZig Zag Linked list \n");
    printList(head);
  
    return (0);
}

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Java

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// Java program to arrange 
// linked list in zigzag fashion 
class GfG
  
/* Link list Node */
static class Node 
    int data; 
    Node next; 
}
static Node head = null;
static int temp = 0;
  
// This function distributes 
// the Node in zigzag fashion 
static void zigZagList(Node head) 
    // If flag is true, then 
    // next node should be greater 
    // in the desired output. 
    boolean flag = true
  
    // Traverse linked list starting from head. 
    Node current = head; 
    while (current != null && current.next != null
    
        if (flag == true) /* "<" relation expected */
        
            /* If we have a situation like A > B > C 
            where A, B and C are consecutive Nodes 
            in list we get A > B < C by swapping B 
            and C */
            if (current.data > current.next.data) 
            {
                    temp = current.data;
                    current.data = current.next.data;
                    current.next.data = temp;
            
        
        else /* ">" relation expected */
        
            /* If we have a situation like A < B < C where 
            A, B and C are consecutive Nodes in list we 
            get A < C > B by swapping B and C */
            if (current.data < current.next.data) 
            {
                temp = current.data;
                current.data = current.next.data;
                current.next.data = temp;
            
        
  
        current = current.next; 
          
        /* flip flag for reverse checking */
        flag = !(flag); 
    
  
/* UTILITY FUNCTIONS */
/* Function to push a Node */
static void push( 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); 
  
    /* move the head to point to the new Node */
    (head) = new_Node; 
  
/* Function to print linked list */
static void printList(Node Node) 
    while (Node != null
    
        System.out.print(Node.data + "->"); 
        Node = Node.next; 
    
    System.out.println("NULL"); 
  
/* Driver code*/
public static void main(String[] args) 
    /* Start with the empty list */
    //Node head = null; 
  
    // create a list 4 -> 3 -> 7 -> 8 -> 6 -> 2 -> 1 
    // answer should be -> 3 7 4 8 2 6 1 
    push(1); 
    push(2); 
    push(6); 
    push(8); 
    push( 7); 
    push( 3); 
    push( 4); 
  
    System.out.println("Given linked list "); 
    printList(head); 
  
    zigZagList(head); 
  
    System.out.println("Zig Zag Linked list "); 
    printList(head); 
}
  
// This code is contributed by 
// Prerna Saini.

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C#

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// C# program to arrange 
// linked list in zigzag fashion 
using System;
  
class GfG 
  
    /* Link list Node */
    class Node 
    
        public int data; 
        public Node next; 
    
    static Node head = null
    static int temp = 0; 
  
    // This function distributes 
    // the Node in zigzag fashion 
    static void zigZagList(Node head) 
    
        // If flag is true, then 
        // next node should be greater 
        // in the desired output. 
        bool flag = true
  
        // Traverse linked list starting from head. 
        Node current = head; 
        while (current != null && current.next != null
        
            if (flag == true) /* "<" relation expected */
            
                /* If we have a situation like A > B > C 
                where A, B and C are consecutive Nodes 
                in list we get A > B < C by swapping B 
                and C */
                if (current != null && 
                    current.next != null && 
                    current.data > current.next.data) 
                
                        temp = current.data; 
                        current.data = current.next.data; 
                        current.next.data = temp; 
                
            
            else /* ">" relation expected */
            
                /* If we have a situation like A < B < C where 
                A, B and C are consecutive Nodes in list we 
                get A < C > B by swapping B and C */
                if (current != null && 
                    current.next != null && 
                    current.data < current.next.data) 
                
                    temp = current.data; 
                    current.data = current.next.data; 
                    current.next.data = temp; 
                
            
  
            current = current.next; 
  
            /* flip flag for reverse checking */
            flag = !(flag); 
        
    
  
    /* UTILITY FUNCTIONS */
    /* Function to push a Node */
    static void push( 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); 
  
        /* move the head to point to the new Node */
        (head) = new_Node; 
    
  
    /* Function to print linked list */
    static void printList(Node Node) 
    
        while (Node != null
        
            Console.Write(Node.data + "->"); 
            Node = Node.next; 
        
        Console.WriteLine("NULL"); 
    
  
    /* Driver code*/
    public static void Main() 
    
        /* Start with the empty list */
        //Node head = null; 
  
        // create a list 4 -> 3 -> 7 -> 8 -> 6 -> 2 -> 1 
        // answer should be -> 3 7 4 8 2 6 1 
        push(1); 
        push(2); 
        push(6); 
        push(8); 
        push( 7); 
        push( 3); 
        push( 4); 
  
        Console.WriteLine("Given linked list "); 
        printList(head); 
  
        zigZagList(head); 
  
        Console.WriteLine("Zig Zag Linked list "); 
        printList(head); 
    
/* This code is contributed PrinciRaj1992 */

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Output :

Given linked list 
4->3->7->8->6->2->1->NULL

Zig Zag Linked list 
3->7->4->8->2->6->1->NULL 

In above code, push function pushes the node at the front of the linked list, the code can be easily modified for pushing node at the end of list. Other thing to note is, swapping of data between two nodes is done by swap by value not swap by links for simplicity, for swap by links technique please see this.

Time complexity : O(n)
Auxiliary Space : O(1)

This article is contributed by Utkarsh Trivedi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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