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
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 << "\nLinked list after insertion: "; display(head); return 0; } |
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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 |
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C#
// C# implementation to insert node // at the middle of the linked list using System; public class LinkedList { static Node head; // head of list /* Node Class */ public class Node { public int data; public Node next; // Constructor to create a new node public 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) { Console.Write(temp.data + " "); temp = temp.next; } } // Driver code public static void Main () { // 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); Console.WriteLine("Linked list before "+ "insertion: "); display(); int x = 3; insertAtMid(x); Console.WriteLine("\nLinked list after"+ " insertion: "); display(); } } /* This code contributed by PrinciRaj1992 */ |
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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; } |
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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 |
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Python3
# Python implementation to insert node # at the middle of the linked list # Node Class class Node : def __init__(self, d): self.data = d self.next = None class LinkedList: # function to insert node at the # middle of the linked list def __init__(self): self.head = None # Function to insert a new node # at the beginning def push(self, new_data): new_node = Node(new_data) new_node.next = self.head self.head = new_node def insertAtMid(self, x): # if list is empty if (self.head == None): self.head = Node(x) else: # get a new node newNode = Node(x) # assign values to the slow # and fast pointers slow = self.head fast = self.head.next while (fast != None and fast.next != None): # 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 def display(self): temp = self.head while (temp != None): print(temp.data, end = " "), temp = temp.next # Driver Code # Creating the list 1.2.4.5 ll = LinkedList() ll.push(5) ll.push(4) ll.push(2) ll.push(1) print("Linked list before insertion: "), ll.display() x = 3ll.insertAtMid(x) print("\nLinked list after insertion: "), ll.display() # This code is contributed by prerna saini |
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C#
// C# implementation to insert node // at the middle of the linked list using System; public class LinkedList { static Node head; // head of list /* Node Class */ class Node { public int data; public Node next; // Constructor to create a new node public 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) { Console.Write(temp.data + " "); temp = temp.next; } } // Driver code 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); Console.WriteLine("Linked list before"+ " insertion: "); display(); int x = 3; insertAtMid(x); Console.WriteLine("\nLinked list after"+ " insertion: "); display(); } } // This code is contributed by Rajput-Ji |
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Output:
Linked list before insertion: 1 2 4 5 Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n)
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