You are given a list, L, and another list, P, containing integers sorted in ascending
order. The operation printLots(L, P) will print the elements in L that are in positions
specified by P. For instance, if P = 1, 3, 4, 6, the elements in positions 1, 3, 4, and 6
in L are printed. Write the procedure printLots(L, P). You may use only the public
Collections API container operations. What is the running time of your procedure?

Explanation:

#include <iostream>

#include <stdlib.h>

#include <cstdio>

#define NULL 0

using namespace std;

struct node

{

   int elements;

   struct node *link;

};

typedef struct node NODE;

NODE* create(NODE* h, int P);

int count(NODE* h);

NODE* sort(NODE* h);

void printLots(NODE* L, NODE* P);

NODE* create(NODE* h, int P)

{

   if (h == NULL)

   {

       h = (NODE*)malloc(sizeof(NODE));

       h->elements = P;

       h->link = NULL;

   }

   else

   {

       h->link = create(h->link, P);

   }

   return (h);

   printf("h");

}

int count(NODE* h)

{

   int count = 0;

   if (h != NULL)

   {

       count++;

       h = h->link;

   }

   return count;

}

NODE* sort(NODE* h)

{

   int i, q, r, temp;

   NODE* htemp = h;

   i = count(h);

   for (q = 0; q < i; q++)

   {

       for (r = 0; r < i - 1; r++)

       {

           if (h->elements > h->link->elements)

           {

               temp = h->elements;

               h->elements = h->link->elements;

               h->link->elements = temp;

           }

           h = h->link;

       }

       h = htemp;

   }

   return htemp;

}

void printLots(NODE* L, NODE* P)

{

   int countL, countP, i, j;

   NODE* temp = L;

   P = sort(P);

   L = sort(L);

   countL = count(L);

   countP = count(P);

   for (i = 0; i < countP; i++)

   {

       if (P->elements > countL)

       {

           printf("The L list does not contain %d elements\n", P->elements);

       }

       else

       {

           for (j = 0; j < P->elements - 1; j++)

           {

               L = L->link;

               printf("%d\n", L->elements);

               L = temp;

           }

       }

       P = P->link;

   }

}

int main()  

{

   int cs, value;

   NODE *L, *P;

   L = NULL;

   P = NULL;

   while (1)

   {

       printf("1.Add to list L\n");

       printf("2.Add to list P\n");

       printf("3.Print List\n");

       printf("4.Clear Screen\n");

       printf("5.Exit\n");

       printf("Enter your choice: ");

       scanf_s("%d", &cs);

       switch (cs)

       {

       case 1: scanf_s("%d", &value);

           L = create(L, value);

           break;

       case 2: scanf_s("%d", &value);

           P = create(P, value);

           break;

       case 3: printLots(L, P);

           break;

       case 4: system("cls");

           break;

       case 5: exit(1);

           break;

       }

   }

}

you go to add your answer type bewteen 5000 and 20 characters and can only have 2 people answer on 1 question.your challanges are based on your questions.