Question

Pull out question 8 on Exercises.

See also: Program 4.16 invert (), Program 4.4 printList ()

Configuring -main ()

1. Create a linked list

2. Function call of 8 (a) and (b)

3. Check the accuracy of the output of r list and l list

4. Repeat steps 1-3 above

Please use c language

void printlist(listPointer first) printf ("The list contains: ) for (; first; first = first→link) printf ("%4d", first-"data); printf("In") Program 4.4: Printing a list the nodes are also in this order. Following the merge, x and y do not exist as indi- vidual lists. Each node initially in x or y is now in z. No additional nodes may be used. What is the time complexity of your algorithm? Let list,-(XI, X2, . . . , xn) and list 2-01, уг, . . . , ym). Write a function to merge the two lists together to obtain the linked list, list 7. ,y22 Xm) İfrn > n. 8. S It is possible to traverse a linked list in both directions (i.e., left to right and res- tricted right-to-left) by reversing the links during the left-to-right traversal. A pos- sible configuration for the list under this scheme is given in Figure 4.10. The vari- able r points to the node currently being examined and I to the node on its left. Note that all nodes to the left of r have their links reversed. (a) Write a function to move r to the right n nodes from a given position (1, r) Write a function to move r to the left n nodes from any given position (l, (b) as listPointer invert (listPointer lead) /invert the list pointed to by lead / listPointer middle,trail; middle-NULL while (lead) t resent trail = middle; middle lead; lead lead-ilink; middle-link trail; return middle; poit Program 4.16: Inverting a singly linked list

Answer 1

#include<stdio.h>
#include<stdlib.h>
  
/* Link list node */
struct Node
{
int data;
struct Node* next;
};
  
// This function rotates a linked list counter-clockwise and
// updates the head. The function assumes that k is smaller
// than size of linked list. It doesn't modify the list if
// k is greater than or equal to size
void rotate(struct Node **head_ref, int k)
{
if (k == 0)
return;
  
// Let us understand the below code for example k = 4 and
// list = 10->20->30->40->50->60.
struct Node* current = *head_ref;
  
// current will either point to kth or NULL after this loop.
// current will point to node 40 in the above example
int count = 1;
while (count < k && current != NULL)
{
current = current->next;
count++;
}
  
// If current is NULL, k is greater than or equal to count
// of nodes in linked list. Don't change the list in this case
if (current == NULL)
return;
  
// current points to kth node. Store it in a variable.
// kthNode points to node 40 in the above example
struct Node *kthNode = current;
  
// current will point to last node after this loop
// current will point to node 60 in the above example
while (current->next != NULL)
current = current->next;
  
// Change next of last node to previous head
// Next of 60 is now changed to node 10
current->next = *head_ref;
  
// Change head to (k+1)th node
// head is now changed to node 50
*head_ref = kthNode->next;
  
// change next of kth node to NULL
// next of 40 is now NULL
kthNode->next = NULL;
}
  
/* UTILITY FUNCTIONS */
/* Function to push a node */
void push (struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node =
(struct Node*) malloc(sizeof(struct Node));
  
/* put in the data */
new_node->data = new_data;
  
/* link the old list off the new node */
new_node->next = (*head_ref);
  
/* move the head to point to the new node */
(*head_ref) = new_node;
}
  
/* Function to print linked list */
void printList(struct Node *node)
{
while (node != NULL)
{
printf("%d ", node->data);
node = node->next;
}
}
  
/* Drier program to test above function*/
int main(void)
{
/* Start with the empty list */
struct Node* head = NULL;
  
// create a list 10->20->30->40->50->60
for (int i = 60; i > 0; i -= 10)
push(&head, i);
  
printf("Given linked list \n");
printList(head);
rotate(&head, 4);
  
printf("\nRotated Linked list \n");
printList(head);
  
return (0);
}