Reverse the order of all nodes at even position in given Linked List
Last Updated :
07 Mar, 2022
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 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
# 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# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(1)