Insert N elements in a Linked List one after other at middle position

Given an array of N elements. The task is to insert the given elements at the middle position in the linked list one after another. Each insert operation should take O(1) time complexity.
Examples:

Input: arr[] = {1, 2, 3, 4, 5} 
Output: 1 -> 3 -> 5 -> 4 -> 2 -> NULL 
1 -> NULL 
1 -> 2 -> NULL 
1 -> 3 -> 2 -> NULL 
1 -> 3 -> 4 -> 2 -> NULL 
1 -> 3 -> 5 -> 4 -> 2 -> NULL
Input: arr[] = {5, 4, 1, 2} 
Output: 5 -> 1 -> 2 -> 4 -> NULL 

Approach: There are two cases: 

1. Number of elements present in the list are less than 2.
2. Number of elements present in the list are more than 2. 
   • The number of elements already present are even say N then the new element is inserted in the middle position that is (N / 2) + 1.
   • The number of elements already present are odd then the new element is inserted next to the current middle element that is (N / 2) + 2.

We take one additional pointer ‘middle’ which stores the address of current middle element and a counter which counts the total number of elements. 
If the elements already present in the linked list are less than 2 then middle will always point to the first position and we insert the new node after the current middle. 
If the elements already present in the linked list are more than 2 then we insert the new node next to the current middle and increment the counter. 
If there are an odd number of elements after insertion then the middle points to the newly inserted node else there is no change in the middle pointer.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <iostream>
using namespace std;

// Node structure
struct Node {

    int value;
    struct Node* next;
};

// Class to represent a node
// of the linked list
class LinkedList {

private:
    struct Node *head, *mid;
    int count;

public:
    LinkedList();
    void insertAtMiddle(int);
    void show();
};

LinkedList::LinkedList()
{
    head = NULL;
    mid = NULL;
    count = 0;
}

// Function to insert a node in
// the middle of the linked list
void LinkedList::insertAtMiddle(int n)
{

    struct Node* temp = new struct Node();
    struct Node* temp1;

    temp->next = NULL;
    temp->value = n;

    // If the number of elements
    // already present are less than 2
    if (count < 2) {
        if (head == NULL) {
            head = temp;
        }
        else {
            temp1 = head;
            temp1->next = temp;
        }
        count++;

        // mid points to first element
        mid = head;
    }

    // If the number of elements already present
    // are greater than 2
    else {

        temp->next = mid->next;
        mid->next = temp;
        count++;

        // If number of elements after insertion
        // are odd
        if (count % 2 != 0) {

            // mid points to the newly
            // inserted node
            mid = mid->next;
        }
    }
}

// Function to print the nodes
// of the linked list
void LinkedList::show()
{

    struct Node* temp;

    temp = head;

    // Initializing temp to head
    // Iterating and printing till
    // The end of linked list
    // That is, till temp is null
    while (temp != NULL) {

        cout << temp->value << " -> ";
        temp = temp->next;
    }
    cout << "NULL";
    cout << endl;
}

// Driver code
int main()
{
    // Elements to be inserted one after another
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);

    LinkedList L1;

    // Insert the elements
    for (int i = 0; i < n; i++)
        L1.insertAtMiddle(arr[i]);

    // Print the nodes of the linked list
    L1.show();

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Node ure
static class Node
{

    int value;
    Node next;
};

// Class to represent a node
// of the linked list
static class LinkedList
{
    Node head, mid;
    int count;


LinkedList()
{
    head = null;
    mid = null;
    count = 0;
}

// Function to insert a node in
// the middle of the linked list
void insertAtMiddle(int n)
{

    Node temp = new Node();
    Node temp1;

    temp.next = null;
    temp.value = n;

    // If the number of elements
    // already present are less than 2
    if (count < 2)
    {
        if (head == null)
        {
            head = temp;
        }
        else
        {
            temp1 = head;
            temp1.next = temp;
        }
        count++;

        // mid points to first element
        mid = head;
    }

    // If the number of elements already present
    // are greater than 2
    else
    {

        temp.next = mid.next;
        mid.next = temp;
        count++;

        // If number of elements after insertion
        // are odd
        if (count % 2 != 0)
        {

            // mid points to the newly
            // inserted node
            mid = mid.next;
        }
    }
}

// Function to print the nodes
// of the linked list
void show()
{

    Node temp;

    temp = head;

    // Initializing temp to head
    // Iterating and printing till
    // The end of linked list
    // That is, till temp is null
    while (temp != null)
    {

        System.out.print( temp.value + " -> ");
        temp = temp.next;
    }
    System.out.print( "null");
    System.out.println();
}

}

// Driver code
public static void main(String args[])
{
    // Elements to be inserted one after another
    int arr[] = { 1, 2, 3, 4, 5 };
    int n = arr.length;

    LinkedList L1=new LinkedList();

    // Insert the elements
    for (int i = 0; i < n; i++)
        L1.insertAtMiddle(arr[i]);

    // Print the nodes of the linked list
    L1.show();
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach

# Node ure
class Node:
    def __init__(self):
        self.value = 0
        self.next = None

# Class to represent a node
# of the linked list
class LinkedList:

    def __init__(self) :
        self.head = None
        self.mid = None
        self.count = 0

    # Function to insert a node in
    # the middle of the linked list
    def insertAtMiddle(self , n):

        temp = Node()
        temp1 = None

        temp.next = None
        temp.value = n

        # If the number of elements
        # already present are less than 2
        if (self.count < 2):

            if (self.head == None) :

                self.head = temp

            else:

                temp1 = self.head
                temp1.next = temp

            self.count = self.count + 1

            # mid points to first element
            self.mid = self.head

        # If the number of elements already present
        # are greater than 2
        else:

            temp.next = self.mid.next
            self.mid.next = temp
            self.count = self.count + 1

            # If number of elements after insertion
            # are odd
            if (self.count % 2 != 0):

                # mid points to the newly
                # inserted node
                self.mid = self.mid.next

    # Function to print the nodes
    # of the linked list
    def show(self):
        temp = None

        temp = self.head

        # Initializing temp to self.head
        # Iterating and printing till
        # The end of linked list
        # That is, till temp is None
        while (temp != None) :

            print( temp.value, end = " -> ")
            temp = temp.next

        print( "None")

# Driver code

# Elements to be inserted one after another
arr = [ 1, 2, 3, 4, 5]
n = len(arr)

L1 = LinkedList()

# Insert the elements
for i in range(n):
    L1.insertAtMiddle(arr[i])

# Print the nodes of the linked list
L1.show()

# This code is contributed by Arnab Kundu

C#

// C# implementation of the approach
using System;

class GFG
{

// Node ure
public class Node
{

    public int value;
    public Node next;
};

// Class to represent a node
// of the linked list
public class LinkedList
{
    public Node head, mid;
    public int count;


public LinkedList()
{
    head = null;
    mid = null;
    count = 0;
}

// Function to insert a node in
// the middle of the linked list
public void insertAtMiddle(int n)
{

    Node temp = new Node();
    Node temp1;

    temp.next = null;
    temp.value = n;

    // If the number of elements
    // already present are less than 2
    if (count < 2)
    {
        if (head == null)
        {
            head = temp;
        }
        else
        {
            temp1 = head;
            temp1.next = temp;
        }
        count++;

        // mid points to first element
        mid = head;
    }

    // If the number of elements already present
    // are greater than 2
    else
    {

        temp.next = mid.next;
        mid.next = temp;
        count++;

        // If number of elements after insertion
        // are odd
        if (count % 2 != 0)
        {

            // mid points to the newly
            // inserted node
            mid = mid.next;
        }
    }
}

// Function to print the nodes
// of the linked list
public void show()
{

    Node temp;

    temp = head;

    // Initializing temp to head
    // Iterating and printing till
    // The end of linked list
    // That is, till temp is null
    while (temp != null)
    {

        Console.Write( temp.value + " -> ");
        temp = temp.next;
    }
    Console.Write( "null");
    Console.WriteLine();
}

}

// Driver code
public static void Main(String []args)
{
    // Elements to be inserted one after another
    int []arr = { 1, 2, 3, 4, 5 };
    int n = arr.Length;

    LinkedList L1=new LinkedList();

    // Insert the elements
    for (int i = 0; i < n; i++)
        L1.insertAtMiddle(arr[i]);

    // Print the nodes of the linked list
    L1.show();
}
}

// This code contributed by Rajput-Ji

Javascript

<script>

      // JavaScript implementation of the approach
      // Node ure
      class Node {
        constructor() {
          this.value = 0;
          this.next = null;
        }
      }

      // Class to represent a node
      // of the linked list
      class LinkedList {
        constructor() {
          this.head = null;
          this.mid = null;
          this.count = 0;
        }

        // Function to insert a node in
        // the middle of the linked list
        insertAtMiddle(n) {
          var temp = new Node();
          var temp1;

          temp.next = null;
          temp.value = n;

          // If the number of elements
          // already present are less than 2
          if (this.count < 2) {
            if (this.head == null) {
              this.head = temp;
            } else {
              temp1 = this.head;
              temp1.next = temp;
            }
            this.count++;

            // mid points to first element
            this.mid = this.head;
          }

          // If the number of elements already present
          // are greater than 2
          else {
            temp.next = this.mid.next;
            this.mid.next = temp;
            this.count++;

            // If number of elements after insertion
            // are odd
            if (this.count % 2 != 0) {
              // mid points to the newly
              // inserted node
              this.mid = this.mid.next;
            }
          }
        }

        // Function to print the nodes
        // of the linked list
        show() {
          var temp;

          temp = this.head;

          // Initializing temp to head
          // Iterating and printing till
          // The end of linked list
          // That is, till temp is null
          while (temp != null) {
            document.write(temp.value + " -> ");
            temp = temp.next;
          }
          document.write("null");
          document.write("<br>");
        }
      }

      // Driver code
      // Elements to be inserted one after another
      var arr = [1, 2, 3, 4, 5];
      var n = arr.length;

      var L1 = new LinkedList();

      // Insert the elements
      for (var i = 0; i < n; i++) L1.insertAtMiddle(arr[i]);

      // Print the nodes of the linked list
      L1.show();

</script>

Output:

1 -> 3 -> 5 -> 4 -> 2 -> NULL

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