C++ Program For Inserting Node In The Middle Of The Linked List
Last Updated :
23 Jul, 2025
Improve
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
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++ 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;
return newNode;
}
// 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 << "
Linked list after insertion: ";
display(head);
return 0;
}
// C++ implementation to insert node at the middle// of the linked listusing namespace std;// structure of a nodestruct Node { int data; Node* next;};// function to create and return a nodeNode* getNode(int data){ // allocating space Node* newNode = (Node*)malloc(sizeof(Node)); // inserting the required data newNode->data = data; newNode->next = NULL; return newNode;}// function to insert node at the middle// of the linked listvoid 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 listvoid display(Node* head){ while (head != NULL) { cout << head->data << " "; head = head->next; }}// Driver program to test aboveint 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 << "Linked list after insertion: "; display(head); return 0;}Output:
Linked list before insertion: 1 2 4 5 Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n)
Space complexity: O(1) since using constant space
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++ 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;
return newNode;
}
// 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 << "
Linked list after insertion: ";
display(head);
return 0;
}
// C++ implementation to insert node at the middle// of the linked listusing namespace std;// structure of a nodestruct Node { int data; Node* next;};// function to create and return a nodeNode* getNode(int data){ // allocating space Node* newNode = (Node*)malloc(sizeof(Node)); // inserting the required data newNode->data = data; newNode->next = NULL; return newNode;}// function to insert node at the middle// of the linked listvoid 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 listvoid display(Node* head){ while (head != NULL) { cout << head->data << " "; head = head->next; }}// Driver program to test aboveint 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 << "Linked list after insertion: "; display(head); return 0;}Output:
Linked list before insertion: 1 2 4 5 Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n)
Space complexity: O(n) where n is size of linked list
Please refer complete article on Insert node into the middle of the linked list for more details!
