Insert node into the middle of the linked list

24 Jun, 2022
Given a linked list containing n nodes. The problem is to insert a new node with data x in 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 the length of the linked list): Find the number of nodes or length of the linked list 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++
Java
Python3
C#
Javascript
// C++ implementation to insert node at the middle// of the linked list#include <bits/stdc++.h>  using 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 << "\nLinked list after insertion: ";    display(head);      return 0;}
// Java implementation to insert node// at the middle of the linked listimport 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
# Python3 implementation to insert node# at the middle of a linked list  # Node classclass Node:      # constructor to create a new node    def __init__(self, data):        self.data = data        self.next = None  # function to insert node at the# middle of linked list given the headdef insertAtMid(head, x):      if(head == None): #if the list is empty        head = Node(x)    else:                  # create a new node for the value        # to be inserted        newNode = Node(x)          ptr = head        length = 0                  # calculate the length of the linked        # list        while(ptr != None):            ptr = ptr.next            length += 1          # 'count' the number of node after which        # the new node has to be inserted        if(length % 2 == 0):            count = length / 2         else:            (length + 1) / 2          ptr = head          # move ptr to the node after which        # the new node has to inserted        while(count > 1):            count -= 1            ptr = ptr.next          # insert the 'newNode' and adjust        # links accordingly        newNode.next = ptr.next        ptr.next = newNode  # function to display the linked listdef display(head):    temp = head    while(temp != None):        print(str(temp.data), end = " ")        temp = temp.next  # Driver Code  # Creating the linked list 1.2.4.5head = Node(1)head.next = Node(2)head.next.next = Node(4)head.next.next.next = Node(5)  print("Linked list before insertion: ", end = "")display(head)  # inserting 3 in the middle of the linked list.x = 3insertAtMid(head, x)  print("\nLinked list after insertion: " , end = "")display(head)  # This code is contributed by Pranav Devarakonda
// 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 */
<script>  // Javascript implementation to insert node// at the middle of the linked list        var head; // head of list      /* Node Class */     class Node {          // Constructor to create a new node        constructor(d) {            this.data = d;            this.next = null;        }    }      // function to insert node at the    // middle of the linked list    function insertAtMid(x) {        // if list is empty        if (head == null)            head = new Node(x);        else {            // get a new node            var newNode = new Node(x);              var ptr = head;            var 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            var 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    function display() {        var temp = head;        while (temp != null) {            document.write(temp.data + " ");            temp = temp.next;        }    }      // Driver program to test above              // Creating the list 1.2.4.5                  head = new Node(1);        head.next = new Node(2);        head.next.next = new Node(4);        head.next.next.next = new Node(5);          document.write("Linked list before "         + "insertion: ");        display();          var x = 3;        insertAtMid(x);          document.write("<br/>Linked list after" +        " insertion: ");        display();  // This code contributed by Rajput-Ji  </script>
Output: 
Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra 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++
Java
Python3
C#
Javascript
// C++ implementation to insert node at the middle// of the linked list#include <bits/stdc++.h>  using 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 << "\nLinked list after insertion: ";    display(head);      return 0;}
// Java implementation to insert node // at the middle of the linked listimport 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
# Python implementation to insert node # at the middle of the linked list   # Node Classclass 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
// C# implementation to insert node // at the middle of the linked listusing 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
<script>  // Javascript implementation to insert node // at the middle of the linked list  var head; // head of list      /* Node Class */     class Node {  // Constructor to create a new nodeconstructor(val) {    this.data = val;    this.next = null;}}      // function to insert node at the    // middle of the linked list    function insertAtMid(x) {        // if list is empty        if (head == null)            head = new Node(x);          else {            // get a new node    var newNode = new Node(x);              // assign values to the slow            // and fast pointers    var slow = head;    var 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      function display() {       var temp = head;        while (temp != null) {            document.write(temp.data + " ");            temp = temp.next;        }    }      // Driver program to test above              // 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);          document.write(        "Linked list before" + " insertion: "        );        display();          var x = 3;        insertAtMid(x);          document.write(        "<br/>Linked list after" + " insertion: "        );        display();  // This code is contributed by todaysgaurav  </script>
Output: 
Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.
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prerna saini
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