Write a function that takes two call-by-reference arguments that are pointer variables that point to the heads of linked lists of values of type int. The two linked lists are assumed to be sorted so that the number at the head is the smallest number, the number in the next node is the next smallest, and so forth. The function returns a pointer to the head of a new linked list that contains all of the nodes in the original two lists.
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- Write a void function that takes a linked list of integers and removes all the duplicate elements from the list. The function will have one call-by-reference parameter that is a pointer to the head of the list. After the function is called, this pointer will point to the head of a linked list that has the nodes that have all the distinct integers. Place your function in a suitable test program.Write a function that gets a linked list of ints, and reverses it. For example - on input 1 -> 2 -> 3 -> 4, after the function finishes the execution, the list becomes 4 -> 3 -> 2 -> 1 - If the list has one element, then it doesn’t change - If the list is empty, then it doesn’t change You may use the data fields in the struct and the functions provided in LL.h and LL.c. // reverses a linked list void LL_reverse(LL_t* list); Test for the Function: void test_q3() { LL_t* lst = LLcreate(); LL_add_to_tail(lst, 1); LL_add_to_tail(lst, 3); LL_add_to_tail(lst, 8); LL_add_to_tail(lst, 4); LL_add_to_tail(lst, 3); LL_reverse(lst); intcorrect[] = {3,4,8,3,1}; inti; node_t* n = lst->head; for(i=0;i<5;i++) { if (n==NULL) { printf("Q3 ERROR: node %d==NULL unexpected\n", i); return; } if (n->data != correct[i]) { printf("Q3 ERROR: node%d->data==%d, expected %d\n", i, n->data, correct[i]); return; } n = n->next; } if (n==NULL) printf("Q3 ok\n"); } Support…Write a function that gets a linked list of ints, and reverses it. For example - on input 1 -> 2 -> 3 -> 4, after the function finishes the execution, the list becomes 4 -> 3 -> 2 -> 1 - If the list has one element, then it doesn’t change - If the list is empty, then it doesn’t change You may use the data fields in the struct and the functions provided in LL.h and LL.c. // reverses a linked list void LL_reverse(LL_t* list); Test for the Function; void test_q3() { LL_t* lst = LLcreate(); LL_add_to_tail(lst, 1); LL_add_to_tail(lst, 3); LL_add_to_tail(lst, 8); LL_add_to_tail(lst, 4); LL_add_to_tail(lst, 3); LL_reverse(lst); intcorrect[] = {3,4,8,3,1}; inti; node_t* n = lst->head; for(i=0;i<5;i++) { if (n==NULL) { printf("Q3 ERROR: node %d==NULL unexpected\n", i); return; } if (n->data != correct[i]) { printf("Q3 ERROR: node%d->data==%d, expected %d\n", i, n->data, correct[i]); return; } n = n->next; } if (n==NULL) printf("Q3 ok\n"); } Support File…
- Consider the following function that takes reference to head of a Doubly Linked List as parameter. Assume that anode of doubly linked list has previous pointer as prev and next pointer as next.void fun(struct node **head_ref){struct node *temp = NULL;struct node *current = *head_ref;while (current != NULL){temp = current->prev;current->prev = current->next;current->next = temp;current = current->prev;}if(temp != NULL )*head_ref = temp->prev;}Assume that reference of head of following doubly linked list is passed to above function 1 <--> 2 <--> 3 <--> 4 <--> 5 <-->6. What should be the modified linked list after the function call? Note: solve as soon as possibleSuppose you have two objects of Doubly Linked List D1 and D2. Each object is representing a different Doubly Linked List. Write a function to concatenate both doubly linked lists. The concatenation function must return the address of head node of concatenated Doubly Linked List, and it must take two Node* parameters. You need to keep in mind all exceptional cases for example what if either one of the doubly linked lists is empty? Perform this task on paper, take a clear picture of solution and paste it in answer section.You are given the pointer to the head node of a linked list and an integer to add to the list. Create a new node with the given integer. Insert this node at the tail of the linked list and return the head node of the linked list formed after inserting this new node. The given head pointer may be null, meaning that the initial list is empty. Function Description Complete the insertNodeAtTail function in the editor below. insertNodeAtTail has the following parameters: SinglyLinkedListNode pointer head: a reference to the head of a list int data: the data value for the node to insert Returns SinglyLinkedListNode pointer: reference to the head of the modified linked list Input Format The first line contains an integer , the number of elements in the linked list.The next lines contain an integer each, the value that needs to be inserted at tail. Constraints Sample Input STDIN Function ----- -------- 5 size of linked list n = 5 141 linked list data values…
- in C++ kth ElementExtend the class linkedListType by adding the following operations:a. Write a function that returns the info of the kth element of the linked list. If no such element exists, terminate the program.b. Write a function that deletes the kth element of the linked list. If no such element exists, terminate the program. Provide the definitions of these functions in the class linkedListType. please, do not copy from any other sources, give me a fresh new code. Thank youC++ Create a generic function add_bookends(ls, e) that adds a copy of element e to the front and back of list ls.C++ Create a generic function average({elements}) that returns the average of all the given elements. For example, average({x, y, z}), returns (x + y + z)/3. The argument is an initializer list that is assumed not to be empty. For the sake of the exercise, don’t use other STL algorithms in the implementation of the function. Write a test driver
- Write a function set_elements() that assigns 1 to the last and fifth elements of a list parameter. Ex: If the input is: One man's trash is another man's treasure then the output is: ['One', "man's", 'trash', 'is', 1, "man's", 1] ''' Your code goes here ''' vals_to_update = input().split() set_elements(vals_to_update) print(vals_to_update)Create a generic function print(ls, n) that prints to standard output the first n elements of list ls. The elements are printed on separate lines. If ls has less than n elements, then the entire list is printed. I attached a picture of what I have so far. I'm having a hard time figuring out how I would make it print a certain number of elements that I specify.Use the following code which adds and prints elements of a simply linked list, and add a member function that will duplicate nonzero value nodes in a list. Call your function properly in main and print the list again. #include <iostream> using namespace std; struct node { int data; node *next; node(int d,node*n=0) { data=d; next=n; } }; class list { node *head; public: list() { head=0; } void print() { for (node *t=head;t!=0;t=t->next) cout<<t->data<<" "; cout<<endl; } void add(int el) { head=new node(el,head); } }; void main() { list l; l.add(0); l.add(7); l.add(1); l.add(0); l.add(3); l.print(); }