Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Reverse the order of all nodes at even position in given Linked List

  • Difficulty Level : Easy
  • Last Updated : 23 Dec, 2021

Given a linked list A[] of N integers, the task is to reverse the order of all integers at an even position.

Examples:

Input: A[] = 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> NULL
Output: 1 6 3 4 5 2
Explanation:  Nodes at even position in the given linked list are 2, 4 and 6. So, after reversing there order, the new linked list will be 1 -> 6 -> 3 -> 4 -> 5 -> 2 -> NULL.

Input:  A[] = 1 -> 5 -> 3->NULL
Output: 1 5 3

Approach: The given problem can be solved by maintaining two linked lists, one list for all odd positioned nodes and another list for all even positioned nodes. Traverse the given linked list which will be considered as the odd list. Hence, for all the nodes at even positions, remove it from the odd list and insert it to the front of the even node list. Since the nodes are added at the front, their order will be reversed. Merge the two lists at alternate positions which is the required answer.

Below is the implementation of the above approach.

C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Node of the linked list
class Node {
public:
    int data;
    Node* next;
};
 
// Function to reverse all the even
// positioned node of given linked list
Node* reverse_even(Node* A)
{
    // Stores the nodes with
    // even positions
    Node* even = NULL;
 
    // Stores the nodes with
    // odd positions
    Node* odd = A;
 
    // If size of list is less that
    // 3, no change is required
    if (!odd || !odd->next || !odd->next->next)
        return odd;
 
    // Loop to traverse the list
    while (odd && odd->next) {
 
        // Store the even positioned
        // node in temp
        Node* temp = odd->next;
        odd->next = temp->next;
 
        // Add the even node to the
        // beginning of even list
        temp->next = even;
 
        // Make temp as new even list
        even = temp;
 
        // Move odd to it's next node
        odd = odd->next;
    }
 
    odd = A;
 
    // Merge the evenlist into
    // odd list alternatively
    while (even) {
 
        // Stores the even's next
        // node in temp
        Node* temp = even->next;
 
        // Link the next odd node
        // to next of even node
        even->next = odd->next;
 
        // Link even to next odd node
        odd->next = even;
 
        // Make new even as temp node
        even = temp;
 
        // Move odd to it's 2nd next node
        odd = odd->next->next;
    }
 
    return A;
}
 
// Function to add a node at
// the beginning of Linked List
void push(Node** head_ref, int new_data)
{
    Node* new_node = new Node();
    new_node->data = new_data;
    new_node->next = (*head_ref);
    (*head_ref) = new_node;
}
 
// Function to print nodes
// in a given linked list
void printList(Node* node)
{
    while (node != NULL) {
        cout << node->data << " ";
        node = node->next;
    }
}
 
// Driver Code
int main()
{
    Node* start = NULL;
 
    push(&start, 6);
    push(&start, 5);
    push(&start, 4);
    push(&start, 3);
    push(&start, 2);
    push(&start, 1);
 
    start = reverse_even(start);
    printList(start);
 
    return 0;
}

Java




// Java program for the above approach
 
import java.util.*;
 
class GFG{
 
// Node of the linked list
static class Node {
 
    int data;
    Node next;
};
static Node start = null;
// Function to reverse all the even
// positioned node of given linked list
static Node reverse_even(Node A)
{
    // Stores the nodes with
    // even positions
    Node even = null;
 
    // Stores the nodes with
    // odd positions
    Node odd = A;
 
    // If size of list is less that
    // 3, no change is required
    if (odd==null || odd.next==null || odd.next.next==null)
        return odd;
 
    // Loop to traverse the list
    while (odd!=null && odd.next!=null) {
 
        // Store the even positioned
        // node in temp
        Node temp = odd.next;
        odd.next = temp.next;
 
        // Add the even node to the
        // beginning of even list
        temp.next = even;
 
        // Make temp as new even list
        even = temp;
 
        // Move odd to it's next node
        odd = odd.next;
    }
 
    odd = A;
 
    // Merge the evenlist into
    // odd list alternatively
    while (even!=null) {
 
        // Stores the even's next
        // node in temp
        Node temp = even.next;
 
        // Link the next odd node
        // to next of even node
        even.next = odd.next;
 
        // Link even to next odd node
        odd.next = even;
 
        // Make new even as temp node
        even = temp;
 
        // Move odd to it's 2nd next node
        odd = odd.next.next;
    }
 
    return A;
}
 
// Function to add a node at
// the beginning of Linked List
static void push(int new_data)
{
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = start;
    start = new_node;
}
 
// Function to print nodes
// in a given linked list
static void printList(Node node)
{
    while (node != null) {
        System.out.print(node.data+ " ");
        node = node.next;
    }
}
 
// Driver Code
public static void main(String[] args)
{
  
 
    push(6);
    push(5);
    push(4);
    push(3);
    push(2);
    push(1);
 
    start = reverse_even(start);
    printList(start);
 
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python program for the above approach
start = None
 
# Node of the linked list
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# Function to reverse all the even
# positioned node of given linked list
def reverse_even(A):
   
    # Stores the nodes with
    # even positions
    even = None
 
    # Stores the nodes with
    # odd positions
    odd = A
 
    # If size of list is less that
    # 3, no change is required
    if (odd == None or odd.next == None or odd.next.next == None):
        return odd
 
    # Loop to traverse the list
    while (odd and odd.next):
 
        # Store the even positioned
        # node in temp
        temp = odd.next
        odd.next = temp.next
 
        # Add the even node to the
        # beginning of even list
        temp.next = even
 
        # Make temp as new even list
        even = temp
 
        # Move odd to it's next node
        odd = odd.next
 
    odd = A
 
    # Merge the evenlist into
    # odd list alternatively
    while (even):
 
        # Stores the even's next
        # node in temp
        temp = even.next
 
        # Link the next odd node
        # to next of even node
        even.next = odd.next
 
        # Link even to next odd node
        odd.next = even
 
        # Make new even as temp node
        even = temp
 
        # Move odd to it's 2nd next node
        odd = odd.next.next
    return A
 
# Function to add a node at
# the beginning of Linked List
def push(new_data):
    global start
    new_node = Node(new_data)
    new_node.next = start
    start = new_node
 
# Function to print nodes
# in a given linked list
def printList(node):
    while (node != None):
        print(node.data, end=" ")
        node = node.next
 
# Driver Code
start = None
 
push(6)
push(5)
push(4)
push(3)
push(2)
push(1)
 
start = reverse_even(start)
printList(start)
 
# This code is contributed by saurabh_jaiswal.

C#




// C# program for the above approach
using System;
public class GFG{
 
// Node of the linked list
class Node {
 
    public int data;
    public Node next;
};
static Node start = null;
   
// Function to reverse all the even
// positioned node of given linked list
static Node reverse_even(Node A)
{
   
    // Stores the nodes with
    // even positions
    Node even = null;
 
    // Stores the nodes with
    // odd positions
    Node odd = A;
 
    // If size of list is less that
    // 3, no change is required
    if (odd==null || odd.next==null || odd.next.next==null)
        return odd;
 
    // Loop to traverse the list
    while (odd!=null && odd.next!=null) {
 
        // Store the even positioned
        // node in temp
        Node temp = odd.next;
        odd.next = temp.next;
 
        // Add the even node to the
        // beginning of even list
        temp.next = even;
 
        // Make temp as new even list
        even = temp;
 
        // Move odd to it's next node
        odd = odd.next;
    }
 
    odd = A;
 
    // Merge the evenlist into
    // odd list alternatively
    while (even!=null) {
 
        // Stores the even's next
        // node in temp
        Node temp = even.next;
 
        // Link the next odd node
        // to next of even node
        even.next = odd.next;
 
        // Link even to next odd node
        odd.next = even;
 
        // Make new even as temp node
        even = temp;
 
        // Move odd to it's 2nd next node
        odd = odd.next.next;
    }
 
    return A;
}
 
// Function to add a node at
// the beginning of Linked List
static void push(int new_data)
{
    Node new_node = new Node();
    new_node.data = new_data;
    new_node.next = start;
    start = new_node;
}
 
// Function to print nodes
// in a given linked list
static void printList(Node node)
{
    while (node != null) {
        Console.Write(node.data+ " ");
        node = node.next;
    }
}
 
// Driver Code
public static void Main(String[] args)
{
    push(6);
    push(5);
    push(4);
    push(3);
    push(2);
    push(1);
 
    start = reverse_even(start);
    printList(start);
 
}
}
 
// This code is contributed by shikhasingrajput

Javascript




<script>
// Javascript program for the above approach
 
// Node of the linked list
class Node {
 
 
    constructor(data) {
        this.data = data;
        this.next = null;
    }
};
 
// Function to reverse all the even
// positioned node of given linked list
function reverse_even(A) {
    // Stores the nodes with
    // even positions
    let even = null;
 
    // Stores the nodes with
    // odd positions
    let odd = A;
 
    // If size of list is less that
    // 3, no change is required
    if (odd == null || odd.next == null || odd.next.next == null)
        return odd;
 
    // Loop to traverse the list
    while (odd && odd.next) {
 
        // Store the even positioned
        // node in temp
        let temp = odd.next;
        odd.next = temp.next;
 
        // Add the even node to the
        // beginning of even list
        temp.next = even;
 
        // Make temp as new even list
        even = temp;
 
        // Move odd to it's next node
        odd = odd.next;
    }
 
    odd = A;
 
    // Merge the evenlist into
    // odd list alternatively
    while (even) {
 
        // Stores the even's next
        // node in temp
        let temp = even.next;
 
        // Link the next odd node
        // to next of even node
        even.next = odd.next;
 
        // Link even to next odd node
        odd.next = even;
 
        // Make new even as temp node
        even = temp;
 
        // Move odd to it's 2nd next node
        odd = odd.next.next;
    }
 
    return A;
}
 
// Function to add a node at
// the beginning of Linked List
function push(new_data) {
    let new_node = new Node();
    new_node.data = new_data;
    new_node.next = start;
    start = new_node;
}
 
// Function to print nodes
// in a given linked list
function printList(node) {
    while (node != null) {
        document.write(node.data + " ");
        node = node.next;
    }
}
 
// Driver Code
let start = null;
 
push(6);
push(5);
push(4);
push(3);
push(2);
push(1);
 
start = reverse_even(start);
printList(start);
 
// This code is contributed by saurabh_jaiswal.
</script>
Output
1 6 3 4 5 2 

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




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!