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Chapter 17: Problem 7

In a(n) __________ list, the last node has a pointer to the first node.

Short Answer

Expert verified
Answer: Circular list

Step by step solution

01

Identify the type of list

In computer programming, there are two types of lists, which are linear lists and circular lists. Linear lists are the ones in which the last element points to NULL or nothing, whereas circular lists have their last element pointing to the first element of the list. In this exercise, we need to identify the type of list with the last node having a pointer to the first node.
02

Answer the question

From the given information, we know that the last node has a pointer to the first node. This type of list is called a circular list because it forms a loop or circle with its elements. Therefore, the answer is: In a(n) circular list, the last node has a pointer to the first node.

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Most popular questions from this chapter

The __________ points to the first node in a linked list.

Using the ListNode structure introduced in this chapter, write a function void printFirst(ListNode *ptr) that prints the value stored in the first node of a list passed to it as parameter. The function should print an error message and terminate the program if the list passed to it is empty.

A data structure that points to an object of the same type as itself is known as a(n) __________ data structure.

Each of the following member functions for performing an operation on a linked list of type NumberList has at least one error. Explain what is wrong and how to fix it. A) NumberList: :printList ( ) while (head) cout \(<<\) head- \(>\) value head \(=\) head- \(>\) next B) NumberList: :printList ( ) ListNode \(\star \mathrm{p}=\) head; while \((\mathrm{p}->\text { next })\) cout \(<\) value \(p=p->n e x t\) C) NumberList: :printList ( ) Listrode \(* p=\) head while \((p)\) cout \(<\) value \(p++;\) D) NumberList: : NumberList ( ) ListNode \(*\) nodePtr, *nextNode; nodePtr = head; while (nodePtr != NULL) nextNode \(=\) nodePtr-next nodePtr- \(>\) next \(=\) NULL \(;\) nodePtr \(=\) nextNode;

Write a function ListNode *removeFirst(ListNode *ptr) that is passed a linked list as parameter, and returns the tail of the list: that is, it removes the first node and returns what is left. The function should deal locate the storage of the removed node. The function returns NULL if the list passed to it is empty.
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