Given an array of N elements. The task is to insert the given elements at the middle position in the linked list one after another. Each insert operation should take O(1) time complexity.
Examples:
Input: arr[] = {1, 2, 3, 4, 5}
Output: 1 -> 3 -> 5 -> 4 -> 2 -> NULL
1 -> NULL
1 -> 2 -> NULL
1 -> 3 -> 2 -> NULL
1 -> 3 -> 4 -> 2 -> NULL
1 -> 3 -> 5 -> 4 -> 2 -> NULLInput: arr[] = {5, 4, 1, 2}
Output: 5 -> 1 -> 2 -> 4 -> NULL
Approach: There are two cases:
- Number of elements present in the list are less than 2.
- Number of elements present in the list are more than 2.
- The number of elements already present are even say N then the new element is inserted in the middle position that is (N / 2) + 1.
- The number of elements already present are odd then the new element is inserted next to the current middle element that is (N / 2) + 2.
We take one additional pointer ‘middle’ which stores the address of current middle element and a counter which counts the total number of elements.
If the elements already present in the linked list are less than 2 then middle will always point to the first position and we insert the new node after the current middle.
If the elements already present in the linked list are more than 2 then we insert the new node next to the current middle and increment the counter.
If there are an odd number of elements after insertion then the middle points to the newly inserted node else there is no change in the middle pointer.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach #include <iostream> using namespace std; // Node structure struct Node { int value; struct Node* next; }; // Class to represent a node // of the linked list class LinkedList { private: struct Node *head, *mid; int count; public: LinkedList(); void insertAtMiddle(int); void show(); }; LinkedList::LinkedList() { head = NULL; mid = NULL; count = 0; } // Function to insert a node in // the middle of the linked list void LinkedList::insertAtMiddle(int n) { struct Node* temp = new struct Node(); struct Node* temp1; temp->next = NULL; temp->value = n; // If the number of elements // already present are less than 2 if (count < 2) { if (head == NULL) { head = temp; } else { temp1 = head; temp1->next = temp; } count++; // mid points to first element mid = head; } // If the number of elements already present // are greater than 2 else { temp->next = mid->next; mid->next = temp; count++; // If number of elements after insertion // are odd if (count % 2 != 0) { // mid points to the newly // inserted node mid = mid->next; } } } // Function to print the nodes // of the linked list void LinkedList::show() { struct Node* temp; temp = head; // Initializing temp to head // Iterating and printing till // The end of linked list // That is, till temp is null while (temp != NULL) { cout << temp->value << " -> "; temp = temp->next; } cout << "NULL"; cout << endl; } // Driver code int main() { // Elements to be inserted one after another int arr[] = { 1, 2, 3, 4, 5 }; int n = sizeof(arr) / sizeof(arr[0]); LinkedList L1; // Insert the elements for (int i = 0; i < n; i++) L1.insertAtMiddle(arr[i]); // Print the nodes of the linked list L1.show(); return 0; } |
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Java
// Java implementation of the approach class GFG { // Node ure static class Node { int value; Node next; }; // Class to represent a node // of the linked list static class LinkedList { Node head, mid; int count; LinkedList() { head = null; mid = null; count = 0; } // Function to insert a node in // the middle of the linked list void insertAtMiddle(int n) { Node temp = new Node(); Node temp1; temp.next = null; temp.value = n; // If the number of elements // already present are less than 2 if (count < 2) { if (head == null) { head = temp; } else { temp1 = head; temp1.next = temp; } count++; // mid points to first element mid = head; } // If the number of elements already present // are greater than 2 else { temp.next = mid.next; mid.next = temp; count++; // If number of elements after insertion // are odd if (count % 2 != 0) { // mid points to the newly // inserted node mid = mid.next; } } } // Function to print the nodes // of the linked list void show() { Node temp; temp = head; // Initializing temp to head // Iterating and printing till // The end of linked list // That is, till temp is null while (temp != null) { System.out.print( temp.value + " -> "); temp = temp.next; } System.out.print( "null"); System.out.println(); } } // Driver code public static void main(String args[]) { // Elements to be inserted one after another int arr[] = { 1, 2, 3, 4, 5 }; int n = arr.length; LinkedList L1=new LinkedList(); // Insert the elements for (int i = 0; i < n; i++) L1.insertAtMiddle(arr[i]); // Print the nodes of the linked list L1.show(); } } // This code is contributed by Arnab Kundu |
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Python3
# Python3 implementation of the approach # Node ure class Node: def __init__(self): self.value = 0 self.next = None # Class to represent a node # of the linked list class LinkedList: def __init__(self) : self.head = None self.mid = None self.count = 0 # Function to insert a node in # the middle of the linked list def insertAtMiddle(self , n): temp = Node() temp1 = None temp.next = None temp.value = n # If the number of elements # already present are less than 2 if (self.count < 2): if (self.head == None) : self.head = temp else: temp1 = self.head temp1.next = temp self.count = self.count + 1 # mid points to first element self.mid = self.head # If the number of elements already present # are greater than 2 else: temp.next = self.mid.next self.mid.next = temp self.count = self.count + 1 # If number of elements after insertion # are odd if (self.count % 2 != 0): # mid points to the newly # inserted node self.mid = self.mid.next # Function to print the nodes # of the linked list def show(self): temp = None temp = self.head # Initializing temp to self.head # Iterating and printing till # The end of linked list # That is, till temp is None while (temp != None) : print( temp.value, end = " -> ") temp = temp.next print( "None") # Driver code # Elements to be inserted one after another arr = [ 1, 2, 3, 4, 5] n = len(arr) L1 = LinkedList() # Insert the elements for i in range(n): L1.insertAtMiddle(arr[i]) # Print the nodes of the linked list L1.show() # This code is contributed by Arnab Kundu |
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C#
// C# implementation of the approach using System; class GFG { // Node ure public class Node { public int value; public Node next; }; // Class to represent a node // of the linked list public class LinkedList { public Node head, mid; public int count; public LinkedList() { head = null; mid = null; count = 0; } // Function to insert a node in // the middle of the linked list public void insertAtMiddle(int n) { Node temp = new Node(); Node temp1; temp.next = null; temp.value = n; // If the number of elements // already present are less than 2 if (count < 2) { if (head == null) { head = temp; } else { temp1 = head; temp1.next = temp; } count++; // mid points to first element mid = head; } // If the number of elements already present // are greater than 2 else { temp.next = mid.next; mid.next = temp; count++; // If number of elements after insertion // are odd if (count % 2 != 0) { // mid points to the newly // inserted node mid = mid.next; } } } // Function to print the nodes // of the linked list public void show() { Node temp; temp = head; // Initializing temp to head // Iterating and printing till // The end of linked list // That is, till temp is null while (temp != null) { Console.Write( temp.value + " -> "); temp = temp.next; } Console.Write( "null"); Console.WriteLine(); } } // Driver code public static void Main(String []args) { // Elements to be inserted one after another int []arr = { 1, 2, 3, 4, 5 }; int n = arr.Length; LinkedList L1=new LinkedList(); // Insert the elements for (int i = 0; i < n; i++) L1.insertAtMiddle(arr[i]); // Print the nodes of the linked list L1.show(); } } // This code contributed by Rajput-Ji |
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Output:
1 -> 3 -> 5 -> 4 -> 2 -> NULL
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