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C++ Program For Inserting Node In The Middle Of The Linked List

  • Last Updated : 11 Jan, 2022

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++




// 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;
}

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;
    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;
}

Output: 

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

Time Complexity: O(n)

Please refer complete article on Insert node into the middle of the linked list for more details!


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