Use the following node definition for this problem. struct NodeInt32 { int32_t value; NodeInt32* next; } Write a function which searches a non-empty linked list for a target value. Its exact signature should be: NodeInt32* find(NodeInt32* head, int32_t target); The function should return the first node whose value equals target. If the target is not found in the list, then the function should return NULL.

Write a display function and delete function in the following code: Declare the libraries Declare the struct Node void create (int A[], int n) { int i; struct Node *t, *last; first = (struct Node *)malloc(sizeof(struct Node)); first->data=A[0]; first->next=NULL; last=first; for(i=1;i<n;i++) { t=(struct Node *)malloc(sizeof(struct Node)); t->data=A[i]; t->next=NULL; last->next=t; last=t; } } Write the Delete Function Write the display function int main() { int A[]={10,20,30,40,50}; create(A,5); Delete(first,4); Display(first); return 0; }

In c++, how do I display the last node in a linked list?

struct insert_at_back_of_sll { // Function takes a constant Book as a parameter, inserts that book at the // back of a singly linked list, and returns nothing. void operator()(const Book& book) { /// TO-DO (3) /// // Write the lines of code to insert "book" at the back of "my_sll". Since // the SLL has no size() function and no tail pointer, you must walk the // list looking for the last node. // // HINT: Do not attempt to insert after "my_sll.end()". // ///// END-T0-DO (3) ||||// } std::forward_list& my_sll; };

Create a user-defined function called duplicates. This function will check to see if a singly linked list contains nodes with the same data stored. If any duplicates are found, return the value 1. Otherwise return 0. You may use the following typedef structure. typedef struct node_s{ int data; struct node_s * nextptr; }node_t;

program Linked List: modify the following program to make a node containing data values of int, char, and string. #include <iostream> using namespace std; struct node { int data; struct Node *next; }; struct Node* head = nullptr;//or Null or 0; void insert(int new_data) { struct Node* new_node=(struct Node*) new(struct Node); new_mode->data=new_data; new_mode->next=head; head=new_node; } void display() { struct Node* ptr; ptr=head; while(ptr ! = NULL) { cout<<ptr->data<<""; ptr=ptr->next; } } int main() { insert{2}; display{}; return0; }

With Head node or not? statement is: if(p->rlink != first) p = p->rlink; O a. It is a Doubly linked list without Head node. O b. It is a Doubly linked list with Head node.