Rearrange a Linked List in Zig-Zag fashion

Difficulty Level : Medium
Last Updated : 05 Feb, 2021

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 the 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++

// 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);
}

Java

// 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.

Python

# Python code to rearrange linked list in zig zac fashion
 
# Node class 
class Node: 
 
    # Constructor to initialize the node object 
    def __init__(self, data): 
        self.data = data 
        self.next = None
 
 
# This function distributes the Node in zigzag fashion
def zigZagList(head):
 
    # If flag is true, then next node should be greater
    # in the desired output.
    flag = True
 
    # Traverse linked list starting from head.
    current = head
    while (current.next != None):
     
        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):
                t = current.data
                current.data = current.next.data
                current.next.data = t
             
         
        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):
                t = current.data
                current.data = current.next.data
                current.next.data = t
             
        current = current.next
        if(flag):
            flag = False # flip flag for reverse checking 
        else:
            flag = True
    return head
 
# function to insert a Node in 
# the linked list at the beginning.
def push(head, k):
 
    tem = Node(0)
    tem.data = k
    tem.next = head
    head = tem
    return head
 
# function to display Node of linked list.
def display( head):
 
    curr = head
    while (curr != None): 
        print( curr.data, "->", end =" ")
        curr = curr.next
     
    print("None")
 
# Driver code
 
head = None
 
# create a list 4 -> 3 -> 7 -> 8 -> 6 -> 2 -> 1
# answer should be -> 3 7 4 8 2 6 1
head = push(head, 1)
head = push(head, 2)
head = push(head, 6)
head = push(head, 8)
head = push(head, 7)
head = push(head, 3)
head = push(head, 4)
 
print("Given linked list \n")
display(head)
 
head = zigZagList(head)
 
print("\nZig Zag Linked list \n")
display(head)
 
# This code is contributed by Arnab Kundu

C#

// 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 */

Output

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

Another Approach:
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.

This can be also be done recursively. The idea remains the same, let us suppose the value of the flag determines the condition we need to check for comparing the current element. So, if the flag is 0 ( or false ) the current element should be smaller than the next and if the flag is 1 ( or true ) then the current element should be greater than the next. If not, swap the values of nodes.

Java

// Java program for the above approach
import java.io.*;
 
// Node class
class Node {
    int data;
    Node next;
    Node(int data) { this.data = data; }
}
 
public class GFG {
 
    private Node head;
 
    // Print Linked List
    public void printLL()
    {
        Node t = head;
        while (t != null) {
            System.out.print(t.data + " ->");
            t = t.next;
        }
        System.out.println();
    }
 
    // Swap both nodes
    public void swap(Node a, Node b)
    {
        if (a == null || b == null)
            return;
        int temp = a.data;
        a.data = b.data;
        b.data = temp;
    }
   
    // Rearrange the linked list
    // in zig zag way
    public Node zigZag(Node node, int flag)
    {
        if (node == null || node.next == null) {
            return node;
        }
        if (flag == 0) {
            if (node.data > node.next.data) {
                swap(node, node.next);
            }
            return zigZag(node.next, 1);
        }
        else {
            if (node.data < node.next.data) {
                swap(node, node.next);
            }
            return zigZag(node.next, 0);
        }
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        GFG lobj = new GFG();
        lobj.head = new Node(11);
        lobj.head.next = new Node(15);
        lobj.head.next.next = new Node(20);
        lobj.head.next.next.next = new Node(5);
        lobj.head.next.next.next.next = new Node(10);
        lobj.printLL();
 
        // 0 means the current element
        // should be smaller than the next
        int flag = 0;
        lobj.zigZag(lobj.head, flag);
        System.out.println("LL in zig zag fashion : ");
        lobj.printLL();
    }
}

Output

11 ->15 ->20 ->5 ->10 ->
LL in zig zag fashion : 
11 ->20 ->5 ->15 ->10 ->

Complexity Analysis:

Time Complexity: O(n).
Traversal of list is done only once and it has ‘n’ elements.
Auxiliary Space: O(1).
No use of extra data structure for storing values.

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

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