Given a singly linked list, find the middle of the linked list. For example, if the given linked list is 1->2->3->4->5 then the output should be 3.
If there are even nodes, then there would be two middle nodes, we need to print the second middle element. For example, if given linked list is 1->2->3->4->5->6 then the output should be 4.
Method 1:
Traverse the whole linked list and count the no. of nodes. Now traverse the list again till count/2 and return the node at count/2.
Java
// Java Program for the above approachimport java.io.*;class GFG { Node head; // Creating a new Node class Node { int data; Node next; public Node(int data) { this.data = data; this.next = null; } } // Function to add a new Node public void pushNode(int data) { Node new_node = new Node(data); new_node.next = head; head = new_node; } // Displaying the elements in the list public void printNode() { Node temp = head; while (temp != null) { System.out.print(temp.data + "->"); temp = temp.next; } System.out.print("Null" + "\n"); } // Finding the length of the list. public int getLen() { int length = 0; Node temp = head; while (temp != null) { length++; temp = temp.next; } return length; } // Printing the middle element of the list. public void printMiddle() { if (head != null) { int length = getLen(); Node temp = head; int middleLength = length / 2; while (middleLength != 0) { temp = temp.next; middleLength--; } System.out.print("The middle element is [" + temp.data + "]"); System.out.print("\n\n"); } } public static void main(String[] args) { GFG list = new GFG(); for (int i = 5; i >= 1; i--) { list.pushNode(i); list.printNode(); list.printMiddle(); } }}// This Code is contributed by Yash Agarwal |
Time Complexity: O(n) where n is no of nodes in linked list
Auxiliary Space: O(1)
Method 2:
Traverse linked list using two pointers. Move one pointer by one and the other pointers by two. When the fast pointer reaches the end slow pointer will reach the middle of the linked list.
Below image shows how printMiddle function works in the code :
Java
// Java program to find middle of// the linked listclass LinkedList{ // Head of linked list Node head; // Linked list node class Node { int data; Node next; Node(int d) { data = d; next = null; } } // Function to print middle of // the linked list void printMiddle() { Node slow_ptr = head; Node fast_ptr = head; if (head != null) { while (fast_ptr != null && fast_ptr.next != null) { fast_ptr = fast_ptr.next.next; slow_ptr = slow_ptr.next; } System.out.println("The middle element is [" + slow_ptr.data + "]"); } } // Inserts a new Node at front of the list. public void push(int new_data) { /* 1 & 2: Allocate the Node & Put in the data*/ Node new_node = new Node(new_data); // 3. Make next of new Node as head new_node.next = head; // 4. Move the head to point to new Node head = new_node; } // This function prints contents of linked list // starting from the given node public void printList() { Node tnode = head; while (tnode != null) { System.out.print(tnode.data + "->"); tnode = tnode.next; } System.out.println("NULL"); } // Driver code public static void main(String [] args) { LinkedList llist = new LinkedList(); for (int i = 5; i > 0; --i) { llist.push(i); llist.printList(); llist.printMiddle(); } }}// This code is contributed by Rajat Mishra |
Output:
5->NULL The middle element is [5] 4->5->NULL The middle element is [5] 3->4->5->NULL The middle element is [4] 2->3->4->5->NULL The middle element is [4] 1->2->3->4->5->NULL The middle element is [3]
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Method 3:
Initialize mid element as head and initialize a counter as 0. Traverse the list from head, while traversing increment the counter and change mid to mid->next whenever the counter is odd. So the mid will move only half of the total length of the list.
Thanks to Narendra Kangralkar for suggesting this method.
Java
// Java program to implement the// above approachclass GFG{ static Node head; // Link list node class Node { int data; Node next; // Constructor public Node(Node next, int data) { this.data = data; this.next = next; } } // Function to get the middle of // the linked list void printMiddle(Node head) { int count = 0; Node mid = head; while (head != null) { // Update mid, when 'count' // is odd number if ((count % 2) == 1) mid = mid.next; ++count; head = head.next; } // If empty list is provided if (mid != null) System.out.println("The middle element is [" + mid.data + "]\n"); } void push(Node head_ref, int new_data) { // Allocate node Node new_node = new Node(head_ref, new_data); // Move the head to point to the new node head = new_node; } // A utility function to print a // given linked list void printList(Node head) { while (head != null) { System.out.print(head.data + "-> "); head = head.next; } System.out.println("null"); } // Driver code public static void main(String[] args) { GFG ll = new GFG(); for(int i = 5; i > 0; i--) { ll.push(head, i); ll.printList(head); ll.printMiddle(head); } }}// This code is contributed by mark_3 |
Output:
5->NULL The middle element is [5] 4->5->NULL The middle element is [5] 3->4->5->NULL The middle element is [4] 2->3->4->5->NULL The middle element is [4] 1->2->3->4->5->NULL The middle element is [3]
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1), no extra space is required, so it is a constant.
Please refer complete article on Find the middle of a given linked list for more details!