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Assume we have a fully implemented singly linked list (similar to the one we learned in the course): 1. Implement a method called mid () which returns the element stored in the middle node as follows: o if the list has odd number of nodes, then output should be the middle element. e.g. if given list is 1->2->3->4->5, then return 3. o If there are even nodes, then there would be two middle nodes; return second middle element. For example, if given list is 1->2->3->4->5->6, then return 4. o If list is empty, return null. Assume you can't use the get(index) method and you have to implement the mid() method from scratch.

Assume we have a fully implemented singly linked list (similar to the one we learned in the course): 1. Implement a method called mid () which returns the element stored in the middle node as follows: o if the list has odd number of nodes, then output should be the middle element. e.g. if given list is 1->2->3->4->5, then return 3. o If there are even nodes, then there would be two middle nodes; return second middle element. For example, if given list is 1->2->3->4->5->6, then return 4. o If list is empty, return null. Assume you can't use the get(index) method and you have to implement the mid() method from scratch.

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Assume we have a fully implemented singly linked list (similar to the one we learned in the course): 1. Implement a method called mid () which returns the element stored in the middle node as follows: o if the list has odd number of nodes, then output should be the middle element. e.g. if given list is 1->2->3->4->5, then return 3. • If there are even nodes, then there would be two middle nodes; return second middle element. For example, if given list is 1->2->3->4->5->6, then return 4. o If list is empty, return null. Assume you can't use the get(index) method and you have to implement the mid() method from scratch. 2. What is the time complexity of your approach in terms of Big-O? public class MyLinkedList<E>{ private Node<E> head, tail; private int size; ... //assume standard methods are implemented here (e.g. isEmpty(), add(), remove(), etc) mplement the mid() method here private class Node<E>{ E element; Node<E> next; public Node(E element) (this.element element;}
Transcribed Image Text:Assume we have a fully implemented singly linked list (similar to the one we learned in the course): 1. Implement a method called mid () which returns the element stored in the middle node as follows: o if the list has odd number of nodes, then output should be the middle element. e.g. if given list is 1->2->3->4->5, then return 3. • If there are even nodes, then there would be two middle nodes; return second middle element. For example, if given list is 1->2->3->4->5->6, then return 4. o If list is empty, return null. Assume you can't use the get(index) method and you have to implement the mid() method from scratch. 2. What is the time complexity of your approach in terms of Big-O? public class MyLinkedList<E>{ private Node<E> head, tail; private int size; ... //assume standard methods are implemented here (e.g. isEmpty(), add(), remove(), etc) mplement the mid() method here private class Node<E>{ E element; Node<E> next; public Node(E element) (this.element element;}
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