Hare-Tortoise Algorithm in Java
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. Also known as Floyd’s Cycle Finding Algorithm.
Below image shows how printMiddle function works in the code :
Below is the implementation of Hare and Tortoise Algorithm:
Java
// Java program to find middle of // the linked list class 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(); } } } |
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]
Complexity of the above method:
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
Alternative Method (Same Concept Hare-Tortoise Algorithm)
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.
Below is the method to finding the middle element of a given Linked List:
Java
// Java program to implement the // above approach class 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); } } } |
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]
Complexity of the above method:
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!
Java Program For Finding The Middle Element Of A Given Linked List
Given a Singly linked list, find the middle of the linked list. If there are even nodes, then there would be two middle nodes, we need to print the second middle element.