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; };