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Reverse even elements in a Linked List

  • Difficulty Level : Medium
  • Last Updated : 09 Jun, 2021

Given a linked list, the task is to reverse the contiguous even elements and print the updated linked list.
 

Input: 1 -> 2 -> 3 -> 3 -> 4 -> 6 -> 8 -> 5 -> NULL 
Output: 1 2 3 3 8 6 4 5 
Initial list: 1 -> 2 -> 3 -> 3 -> 4 -> 6 -> 8 -> 5 -> NULL 
Reversed list: 1 -> 2 -> 3 -> 3 -> 8 -> 6 -> 4 -> 5 -> NULL
Input: 11 -> 32 -> 7 -> NULL 
Output: 11 32 7 
 

 

Approach: Reversing the contiguous even elements will not take place when: 
 

  1. The node’s value is odd.
  2. The node’s value is even but adjacent values are odd.

In rest of the cases, the continuous block of even nodes can be reversed.
Below is the implementation of the above approach: 
 



C++




// C++ implementation of the approach
#include <iostream>
using namespace std;
 
// Structure of a node of the linked list
struct node {
    int data;
    struct node* next;
};
 
// Function to create a new node
struct node* newNode(int d)
{
    struct node* newnode = new node();
    newnode->data = d;
    newnode->next = NULL;
    return newnode;
}
 
// Recursive function to reverse the consecutive
// even nodes of the linked list
struct node* reverse(struct node* head, struct node* prev)
{
 
    // Base case
    if (head == NULL)
        return NULL;
 
    struct node* temp;
    struct node* curr;
    curr = head;
 
    // Reversing nodes until curr node's value
    // turn odd or Linked list is fully traversed
    while (curr != NULL && curr->data % 2 == 0) {
        temp = curr->next;
        curr->next = prev;
        prev = curr;
        curr = temp;
    }
 
    // If elements were reversed then head
    // pointer needs to be changed
    if (curr != head) {
 
        head->next = curr;
 
        // Recur for the remaining linked list
        curr = reverse(curr, NULL);
        return prev;
    }
 
    // Simply iterate over the odd value nodes
    else {
        head->next = reverse(head->next, head);
        return head;
    }
}
 
// Utility function to print the
// contents of the linked list
void printLinkedList(struct node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 3, 4, 6, 8, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    struct node* head = NULL;
    struct node* p;
 
    // Constructing linked list
    for (int i = 0; i < n; i++) {
 
        if (head == NULL) {
            p = newNode(arr[i]);
            head = p;
            continue;
        }
        p->next = newNode(arr[i]);
        p = p->next;
    }
 
    // Head of the updated linked list
    head = reverse(head, NULL);
 
    // Printing the reversed linked list
    printLinkedList(head);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Structure of a node of the linked list
static class node
{
    int data;
    node next;
};
 
// Function to create a new node
static node newNode(int d)
{
    node newnode = new node();
    newnode.data = d;
    newnode.next = null;
    return newnode;
}
 
// Recursive function to reverse the consecutive
// even nodes of the linked list
static node reverse(node head, node prev)
{
 
    // Base case
    if (head == null)
        return null;
 
    node temp;
    node curr;
    curr = head;
 
    // Reversing nodes until curr node's value
    // turn odd or Linked list is fully traversed
    while (curr != null && curr.data % 2 == 0)
    {
        temp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = temp;
    }
 
    // If elements were reversed then head
    // pointer needs to be changed
    if (curr != head)
    {
        head.next = curr;
 
        // Recur for the remaining linked list
        curr = reverse(curr, null);
        return prev;
    }
 
    // Simply iterate over the odd value nodes
    else
    {
        head.next = reverse(head.next, head);
        return head;
    }
}
 
// Utility function to print the
// contents of the linked list
static void printLinkedList(node head)
{
    while (head != null)
    {
        System.out.print(head.data + " ");
        head = head.next;
    }
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 3, 4, 6, 8, 5 };
    int n = arr.length;
 
    node head = null;
    node p = new node();
 
    // Constructing linked list
    for (int i = 0; i < n; i++)
    {
        if (head == null)
        {
            p = newNode(arr[i]);
            head = p;
            continue;
        }
        p.next = newNode(arr[i]);
        p = p.next;
    }
 
    // Head of the updated linked list
    head = reverse(head, null);
 
    // Printing the reversed linked list
    printLinkedList(head);
}
}
 
// This code is contributed by PrinciRaj1992

Python 


# Python implementation of the approach 

# Node of a linked list 
class Node: 
    def __init__(self, next = None, data = None): 
        self.next = next
        self.data = data 

# Function to create a new node
def newNode( d):

    newnode = Node()
    newnode.data = d
    newnode.next = None
    return newnode

# Recursive function to reverse the consecutive
# even nodes of the linked list
def reverse(head, prev):

    # Base case
    if (head == None):
        return None

    temp = None
    curr = None
    curr = head

    # Reversing nodes until curr node's value
    # turn odd or Linked list is fully traversed
    while (curr != None and curr.data % 2 == 0) :
        temp = curr.next
        curr.next = prev
        prev = curr
        curr = temp

    # If elements were reversed then head
    # pointer needs to be changed
    if (curr != head) :
    
        head.next = curr

        # Recur for the remaining linked list
        curr = reverse(curr, None)
        return prev
    
    # Simply iterate over the odd value nodes
    else:
    
        head.next = reverse(head.next, head)
        return head
    
# Utility function to print the
# contents of the linked list
def printLinkedList(head):

    while (head != None):
    
        print(head.data ,end= " ")
        head = head.next
    
# Driver code
arr = [ 1, 2, 3, 3, 4, 6, 8, 5 ]
n = len(arr)

head = None
p = Node()

i = 0

# Constructing linked list
while ( i < n ):
    
    if (head == None):
        
        p = newNode(arr[i])
        head = p
    else:
        p.next = newNode(arr[i])
        p = p.next
    i = i + 1;

# Head of the updated linked list
head = reverse(head, None)

# Printing the reversed linked list
printLinkedList(head)

# This code is contributed by Arnab Kundu


C#




// C# implementation of the above approach
using System;
 
class GFG
{
 
// Structure of a node of the linked list
public class node
{
    public int data;
    public node next;
};
 
// Function to create a new node
static node newNode(int d)
{
    node newnode = new node();
    newnode.data = d;
    newnode.next = null;
    return newnode;
}
 
// Recursive function to reverse the consecutive
// even nodes of the linked list
static node reverse(node head, node prev)
{
 
    // Base case
    if (head == null)
        return null;
 
    node temp;
    node curr;
    curr = head;
 
    // Reversing nodes until curr node's value
    // turn odd or Linked list is fully traversed
    while (curr != null && curr.data % 2 == 0)
    {
        temp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = temp;
    }
 
    // If elements were reversed then head
    // pointer needs to be changed
    if (curr != head)
    {
        head.next = curr;
 
        // Recur for the remaining linked list
        curr = reverse(curr, null);
        return prev;
    }
 
    // Simply iterate over the odd value nodes
    else
    {
        head.next = reverse(head.next, head);
        return head;
    }
}
 
// Utility function to print the
// contents of the linked list
static void printLinkedList(node head)
{
    while (head != null)
    {
        Console.Write(head.data + " ");
        head = head.next;
    }
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 2, 3, 3, 4, 6, 8, 5 };
    int n = arr.Length;
 
    node head = null;
    node p = new node();
 
    // Constructing linked list
    for (int i = 0; i < n; i++)
    {
        if (head == null)
        {
            p = newNode(arr[i]);
            head = p;
            continue;
        }
        p.next = newNode(arr[i]);
        p = p.next;
    }
 
    // Head of the updated linked list
    head = reverse(head, null);
 
    // Printing the reversed linked list
    printLinkedList(head);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// Javascript implementation of the approach
 
// Structure of a node of the linked list
 
class node {
        constructor() {
        this.data = 0;
        this.next = null;
            }
}
 
// Function to create a new node
function newNode( d)
{
    var newnode = new node();
    newnode.data = d;
    newnode.next = null;
    return newnode;
}
 
// Recursive function to reverse the consecutive
// even nodes of the linked list
function reverse(head, prev)
{
 
    // Base case
    if (head == null)
        return null;
 
    var temp;
    var curr;
    curr = head;
 
    // Reversing nodes until curr node's value
    // turn odd or Linked list is fully traversed
    while (curr != null && curr.data % 2 == 0)
    {
        temp = curr.next;
        curr.next = prev;
        prev = curr;
        curr = temp;
    }
 
    // If elements were reversed then head
    // pointer needs to be changed
    if (curr != head)
    {
        head.next = curr;
 
        // Recur for the remaining linked list
        curr = reverse(curr, null);
        return prev;
    }
 
    // Simply iterate over the odd value nodes
    else
    {
        head.next = reverse(head.next, head);
        return head;
    }
}
 
// Utility function to print the
// contents of the linked list
function printLinkedList( head)
{
    while (head != null)
    {
        document.write(head.data + " ");
        head = head.next;
    }
}
 
 
// Driver Code
 
let arr = [ 1, 2, 3, 3, 4, 6, 8, 5 ];
let n = arr.length;
 
var head = null;
var p = new node();
 
// Constructing linked list
for (let i = 0; i < n; i++)
{
    if (head == null)
    {
        p = newNode(arr[i]);
        head = p;
        continue;
    }
    p.next = newNode(arr[i]);
    p = p.next;
}
 
// Head of the updated linked list
head = reverse(head, null);
 
// Printing the reversed linked list
printLinkedList(head);
  
// This code is contributed by jana_sayantan.
</script>
Output: 
1 2 3 3 8 6 4 5

 

Time Complexity: O(N) 
Space Complexity: O(1)
 

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