Reverse even elements in a Linked List

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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Output:

1 2 3 3 8 6 4 5

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



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Improved By : princiraj1992