# Insert node into the middle of the linked list

Given a linked list containing n nodes. The problem is to insert a new node with data x at the middle of the list. If n is even, then insert the new node after the (n/2)th node, else insert the new node after the (n+1)/2th node.

Examples:

```Input : list: 1->2->4->5
x = 3
Output : 1->2->3->4->5

Input : list: 5->10->4->32->16
x = 41
Output : 5->10->4->41->32->16
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1(Using length of the linked list):
Find the number of nodes or length of the linked using one traversal. Let it be len. Calculate c = (len/2), if len is even, else c = (len+1)/2, if len is odd. Traverse again the first c nodes and insert the new node after the cth node.

## C++

```// C++ implementation to insert node at the middle
// of the linked list
#include <bits/stdc++.h>

using namespace std;

// structure of a node
struct Node {
int data;
Node* next;
};

// function to create and return a node
Node* getNode(int data)
{
// allocating space
Node* newNode = (Node*)malloc(sizeof(Node));

// inserting the required data
newNode->data = data;
newNode->next = NULL;
}

// function to insert node at the middle
// of the linked list
void insertAtMid(Node** head_ref, int x)
{
// if list is empty
if (*head_ref == NULL)
*head_ref = getNode(x);
else {

// get a new node
Node* newNode = getNode(x);

Node* ptr = *head_ref;
int len = 0;

// calculate length of the linked list
//, i.e, the number of nodes
while (ptr != NULL) {
len++;
ptr = ptr->next;
}

// 'count' the number of nodes after which
//  the new node is to be inserted
int count = ((len % 2) == 0) ? (len / 2) :
(len + 1) / 2;
ptr = *head_ref;

// 'ptr' points to the node after which
// the new node is to be inserted
while (count-- > 1)
ptr = ptr->next;

// insert the 'newNode' and adjust the
// required links
newNode->next = ptr->next;
ptr->next = newNode;
}
}

// function to display the linked list
void display(Node* head)
{
while (head != NULL) {
cout << head->data << " ";
head = head->next;
}
}

// Driver program to test above
int main()
{
// Creating the list 1->2->4->5
Node* head = NULL;
head = getNode(1);
head->next = getNode(2);
head->next->next = getNode(4);
head->next->next->next = getNode(5);

cout << "Linked list before insertion: ";
display(head);

int x = 3;
insertAtMid(&head, x);

cout << "\nLinked list after insertion: ";
display(head);

return 0;
}
```

## Java

```// Java implementation to insert node
// at the middle of the linked list
import java.util.*;
import java.lang.*;
import java.io.*;

class LinkedList
{
static Node head; // head of list

/* Node Class */
static class Node {
int data;
Node next;

// Constructor to create a new node
Node(int d) {
data = d;
next = null;
}
}

// function to insert node at the
// middle of the linked list
static void insertAtMid(int x)
{
// if list is empty
if (head == null)
head = new Node(x);
else {
// get a new node
Node newNode = new Node(x);

Node ptr = head;
int len = 0;

// calculate length of the linked list
//, i.e, the number of nodes
while (ptr != null) {
len++;
ptr = ptr.next;
}

// 'count' the number of nodes after which
// the new node is to be inserted
int count = ((len % 2) == 0) ? (len / 2) :
(len + 1) / 2;
ptr = head;

// 'ptr' points to the node after which
// the new node is to be inserted
while (count-- > 1)
ptr = ptr.next;

// insert the 'newNode' and adjust
// the required links
newNode.next = ptr.next;
ptr.next = newNode;
}
}

// function to display the linked list
static void display()
{
Node temp = head;
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.next;
}
}

// Driver program to test above
public static void main (String[] args)
{
// Creating the list 1.2.4.5
head = null;
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(4);
head.next.next.next = new Node(5);

System.out.println("Linked list before "+
"insertion: ");
display();

int x = 3;
insertAtMid(x);

System.out.println("\nLinked list after"+
" insertion: ");
display();
}
}

// This article is contributed by Chhavi
```

Output:

```Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
```

Time Complexity: O(n)

Method 2(Using two pointers):
Based on the tortoise and hare algorithm which uses two pointers, one known as slow and the other known as fast. This algorithm helps in finding the middle node of the linked list. It is explained in the front and black split procedure of this post. Now, you can insert the new node after the middle node obtained from the above process. This approach requires only a single traversal of the list.

## C++

```// C++ implementation to insert node at the middle
// of the linked list
#include <bits/stdc++.h>

using namespace std;

// structure of a node
struct Node {
int data;
Node* next;
};

// function to create and return a node
Node* getNode(int data)
{
// allocating space
Node* newNode = (Node*)malloc(sizeof(Node));

// inserting the required data
newNode->data = data;
newNode->next = NULL;
}

// function to insert node at the middle
// of the linked list
void insertAtMid(Node** head_ref, int x)
{
// if list is empty
if (*head_ref == NULL)
*head_ref = getNode(x);

else {
// get a new node
Node* newNode = getNode(x);

// assign values to the slow and fast
// pointers
Node* slow = *head_ref;
Node* fast = (*head_ref)->next;

while (fast && fast->next) {

// move slow pointer to next node
slow = slow->next;

// move fast pointer two nodes at a time
fast = fast->next->next;
}

// insert the 'newNode' and adjust the
// required links
newNode->next = slow->next;
slow->next = newNode;
}
}

// function to display the linked list
void display(Node* head)
{
while (head != NULL) {
cout << head->data << " ";
head = head->next;
}
}

// Driver program to test above
int main()
{
// Creating the list 1->2->4->5
Node* head = NULL;
head = getNode(1);
head->next = getNode(2);
head->next->next = getNode(4);
head->next->next->next = getNode(5);

cout << "Linked list before insertion: ";
display(head);

int x = 3;
insertAtMid(&head, x);

cout << "\nLinked list after insertion: ";
display(head);

return 0;
}
```

## Java

```// Java implementation to insert node
// at the middle of the linked list
import java.util.*;
import java.lang.*;
import java.io.*;

class LinkedList
{
static Node head; // head of list

/* Node Class */
static class Node {
int data;
Node next;

// Constructor to create a new node
Node(int d) {
data = d;
next = null;
}
}

// function to insert node at the
// middle of the linked list
static void insertAtMid(int x)
{
// if list is empty
if (head == null)
head = new Node(x);

else {
// get a new node
Node newNode = new Node(x);

// assign values to the slow
// and fast pointers
Node slow = head;
Node fast = head.next;

while (fast != null && fast.next
!= null)
{
// move slow pointer to next node
slow = slow.next;

// move fast pointer two nodes
// at a time
fast = fast.next.next;
}

// insert the 'newNode' and adjust
// the required links
newNode.next = slow.next;
slow.next = newNode;
}
}

// function to display the linked list
static void display()
{
Node temp = head;
while (temp != null)
{
System.out.print(temp.data + " ");
temp = temp.next;
}
}

// Driver program to test above
public static void main (String[] args)
{
// Creating the list 1.2.4.5
head = null;
head = new Node(1);
head.next = new Node(2);
head.next.next = new Node(4);
head.next.next.next = new Node(5);

System.out.println("Linked list before"+
" insertion: ");
display();

int x = 3;
insertAtMid(x);

System.out.println("\nLinked list after"+
" insertion: ");
display();
}
}

// This article is contributed by Chhavi
```

Output:

```Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
```

Time Complexity: O(n)

This article is contributed by Ayush Jauhari. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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