Given a singly linked list, find middle of the linked list. For example, if given linked list is 1->2->3->4->5 then output should be 3.
If there are even nodes, then there would be two middle nodes, we need to print second middle element. For example, if given linked list is 1->2->3->4->5->6 then output should be 4.
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.
// 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 |
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.
// 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 |
Output:
Linked list before insertion: 1 2 4 5
Linked list after insertion: 1 2 3 4 5
Time Complexity: O(n)
