Question

I don't know how to do the clear function, which mean delete the nodes between n1 and n2

# include using namespace sta; General instructions: Please modify this file and submit the modified one via svn and then websubmission (under the entry "pracExam"). No design documents are needed. struct Node int data; Node* next; /Task 1: Implement the following function for adding a front of the given list. new node to the input Node head: a head pointer to a list int val: an integer that represent the value of the new node return the head pointer of the new list after the insertion Example add a node of value 9 before HEAD-1->2->3-NULL after HEAD->9->1->2-3-NULI Nodeadd (Node* head, int val) I / Task 2 Implement the following function for printing the elements in odd position of the given list. Assume the head node is defined to be in position 1. input Node head: a head pointer to a list return void output The elements in odd position of the given 1ist HEAD->1->2-3-9-12-NULL Example: input output HEAD-1->NA-3-NA->12->NULL void printodd (Node head) /Task 3: Implement the following function for deleting the nodes in between Node nl and Node n2. input Nodes nl a pointer to a node in a 1ist N ae3" n2: a pointer to a node in the зате її3t return void Example: input l points to the 1st node n2 points to the 4th node before HEAD-1->2->3->9-12-NULI after HEAD->1->9-12-NULT void clear (Node nl, Noden2) // You are not supposed to change the main function int main) Node* head = NULL; Node *p4, p7; int test cin >> test; for (int i 1; i 9-8-7->4->3->2-1-NULL //You can uncomment this line to test. //the output should be HEAD->9-NA--NA->3->NA-1->NULL //Please remember to comment this line out before submitting //printodd (head); head add (head, 12) 15); head add (head, = // at this point, the list is: HEAD->15->12->9->8->7->->3->2->1-NULL clear (head, p7) // between p7 and head, there are 3 nodes (12, 9 and 3). Remove them all. // The resulting 1ist is HEAD-15->7-4-3-2-1-ULL printodd (head) /7 the output should be: HEAD->15-»NA->4-»NA->2->NA-NULL // free all nodes Node* tmp; while (head) tmp head; head head->next ; delete tmp; return 0;

Answer 1

This is the code to perform the clear() function:

void clearList (Node *nl, Node n2) 34 Node traverse nl->next; 35 while (traversen2) 36 n1->next traverse->next ; = 37 traverse traverse->neXt 38 39

I will explain how it works now

suppose we have the linked list as 9->8->7->6->5->4->3->2->1, and we need to delete all elements between nodes that point to 9 and 3:

let us see how we delete a node from a linked list:

• a->b->c <---- initial list
• we connect a to c and remove b;
• suppose ponter1 points to a and pointer2 points to b, then we do pointer1->next = pointer2->next; which removes the node b.

this is what we keep doing from n1 to n2.

9->8->7->6->5->4->3->2->1 <--------- initial list
9->7->6->5->4->3->2->1 <--------- deleted 8
9->6->5->4->3->2->1 <--------- deleted 7
9->5->4->3->2->1 <---------deleted 6
9->4->3->2->1 <--------- deleted 5
9->3->2->1 < ------ deleted 4 to get final list

Since you have not provides the test code i have only provided the function you needed. For any further assistance, please comment and if the answer helped please upvote