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

Last Updated : 10 Jul, 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)

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++

// 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)

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!

C++ Program For Inserting A Node In A Linked List

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

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