Important note: This is in C++ format. before posting your solution please make
ID: 3695735 • Letter: I
Question
Important note: This is in C++ format. before posting your solution please make sure the program compiles and run, so that it doesn't waste the question. Also, I would appreciate it if the some parts were commented please, since i am really lost with this. I would truly appreciate the help you can give in this.
Here is the question:
Again this in C++ format. this is based on linked list. Use the linked list operation to write a program to input data for staff member name and salary values (since the number of items is unidentified, use a sentinel value). Calculate the average of salary values and display it. Sort the staff member names in alphabetical order using any sorting method.
Explanation / Answer
#include<iostream>
#include<cstdlib>
using namespace std;
/* Link list node */
struct node
{
int data;
string name;
struct node* next;
};
/* function prototypes */
struct node* SortedMerge(struct node* a, struct node* b);
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef);
/* sorts the linked list by changing next pointers (not data) */
void MergeSort(struct node** headRef)
{
struct node* head = *headRef;
struct node* a;
struct node* b;
/* Base case -- length 0 or 1 */
if ((head == NULL) || (head->next == NULL))
{
return;
}
/* Split head into 'a' and 'b' sublists */
FrontBackSplit(head, &a, &b);
/* Recursively sort the sublists */
MergeSort(&a);
MergeSort(&b);
/* answer = merge the two sorted lists together */
*headRef = SortedMerge(a, b);
}
/* See http://geeksforgeeks.org/?p=3622 for details of this
function */
struct node* SortedMerge(struct node* a, struct node* b)
{
struct node* result = NULL;
/* Base cases */
if (a == NULL)
return(b);
else if (b==NULL)
return(a);
/* Pick either a or b, and recur */
if ((a->name).compare(b->name) <= 0)
{
result = a;
result->next = SortedMerge(a->next, b);
}
else
{
result = b;
result->next = SortedMerge(a, b->next);
}
return(result);
}
/* UTILITY FUNCTIONS */
/* Split the nodes of the given list into front and back halves,
and return the two lists using the reference parameters.
If the length is odd, the extra node should go in the front list.
Uses the fast/slow pointer strategy. */
void FrontBackSplit(struct node* source,
struct node** frontRef, struct node** backRef)
{
struct node* fast;
struct node* slow;
if (source==NULL || source->next==NULL)
{
/* length < 2 cases */
*frontRef = source;
*backRef = NULL;
}
else
{
slow = source;
fast = source->next;
/* Advance 'fast' two nodes, and advance 'slow' one node */
while (fast != NULL)
{
fast = fast->next;
if (fast != NULL)
{
slow = slow->next;
fast = fast->next;
}
}
/* 'slow' is before the midpoint in the list, so split it in two
at that point. */
*frontRef = source;
*backRef = slow->next;
slow->next = NULL;
}
}
/* Function to print nodes in a given linked list */
void printList(struct node *node)
{
while(node!=NULL)
{
cout<<node->name<<" "<<node->data<<endl;
node = node->next;
}
cout<<endl;
}
/* Function to insert a node at the beginging of the linked list */
void push(struct node** head_ref, int salary, string name)
{
/* allocate node */
struct node* new_node = new node;
/* put in the data */
new_node->data = salary;
new_node->name = name;
/* 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;
}
/* Drier program to test above functions*/
int main()
{
/* Start with the empty list */
//struct node* res = NULL;
struct node* a = NULL;
/* Let us create a unsorted linked lists to test the functions
Created lists shall be a: 2->3->20->5->10->15 */
push(&a, 15, "Pk");
push(&a, 10, "Mk");
push(&a, 5, "AK");
push(&a, 20, "VK");
push(&a, 3,"RK");
push(&a, 2, "LK");
/* Sort the above created Linked List */
MergeSort(&a);
cout<<" Sorted Linked List is: ";
printList(a);
return 0;
}
/*
Output:
Sorted Linked List is:
AK 5
LK 2
Mk 10
Pk 15
RK 3
VK 20
*/