Sort a linked list in C++ - c++

Question

I'm trying to wrap my head around how to write an algorithm to sort a linked list, but I'm having a hard time coming up with something that will work. What we need to do have a linked list that contains a name in a string, and an int for hours. After displaying the list and the sum of the hours, we then have to sort the list in ascending order by the hours in a queue. I have the list and all it's functioned stored in a class object, as you will see. I cleared the whole function of what I had in hopes of coming up with a fresh idea but nothing is coming to mind. I initially was going to create a second linked list that had the sorted list, but then I began to wonder if it was possible to sort it within the same list. Here is my code as of posting.
#include <iostream>
#include <ctime>
using namespace std;
// OrderedLL class
template<class T>
class OrderedLL
{
private:
struct NODE
{
string sName;
int sHours;
NODE *next;
};
NODE *list;
NODE *rear;
public:
// Constructor
OrderedLL () { list = rear = NULL;}
// Insert item x -------------------------------------
void Insert(string x, int y)
{
NODE *r;
// Create a new node
r = new(NODE); r->sName = x; r->sHours = y;
r->next = NULL;
// Inserts the item into the list
r->next = list;
list = r;
}
// Display the linked list --------------------------
void display()
{ NODE *p = list;
while( p != NULL)
{ cout << p->sName << "/" << p->sHours << "-->"; p = p->next;}
cout << "NULL\n";
}
// Delete x from the linked list --------------------
void DeleteNode(T x)
{
NODE *p = list, *r = list;
while( p->info != x) {r=p; p=p->next; }
if( p == list)
{ // delete the first node
list = p->next; delete(p);
}
else
{ r->next = p->next; delete(p);}
}
// Sort by hours ------------------------------------
void sortHours()
{
NODE *p, *q;
}
// Display the total hours --------------------------
friend T totHours(OrderedLL LL)
{
NODE *p;
int total = 0;
p = LL.list;
while(p != NULL)
{
total += p->sHours;
p = p->next;
}
cout << "Total spending time = " << total << endl;
}
}; // end of OrderedLL class
int main(void)
{
// Declare variables
time_t a;
OrderedLL<int> unsortedLL;
OrderedLL<int> sortedLL;
int inHours;
string inName;
// Displays the current time and date
time(&a);
cout << "Today is " << ctime(&a) << endl;
// Asks the user to enter a name and hours 5 times, inserting each entry
// into the queue
for(int i = 0; i < 5; i++)
{
cout << "Enter name and Time: ";
cin >> inName >> inHours;
unsortedLL.Insert(inName, inHours);
}
// Displays the unsorted list
cout << "\nWaiting List-->";
unsortedLL.display();
totHours(unsortedLL);
// Calls for the function to sort the list into a queue by hours
unsortedLL.sortHours();
unsortedLL.display();
return 0;
} // End of "main"
As always thanks to anyone who can help

Answer 1

Try sorting the linked-list like you are sorting an integer array. Instead of swapping the nodes, swap the contents inside the nodes.

Answer 2

If you don't care about efficiency you can use any sorting algorithm. What's different between linked lists and arrays is that the swap operation, to swap the position of two elements while sorting, will be a lot slower because it has to run through the links of the list.
An O(n²) bubble sort is easy to implement with a linked list since it only swaps an element with its neighbor.
If you care about efficiency you can look into implementing the merge sort algorithm even if it's a little more complicated.

Answer 3

You should insert like this
void Insert(string x, int y)
{
NODE *r;NODE *temp;
// Create a new node
r = new NODE; r->sName = x; r->sHours = y;r->next = NULL;
if(list==null)//check if list is empty
{
list=r;//insert the node r in the list
}
else
{
temp=list;
while(temp->next!=null)temp=temp->next;//reach to the end of the list
temp->next=r;//insert it at the end of the list
}
}
No need of rear pointer..just check if list->next is null,if yes you are at the end
and your sorthour function should be
void sortHours()
{
for(NODE* n=list;n->next!=null;n=n->next)//get each of the node in list 1 by 1 except the last one i.e. n
{
for(NODE* n1=n->next;n1!=null;n1=n1->next)//compare the list n node with all the nodes that follow it i.e.n1
{
if(n->sHours > n1->sHours)//if one of the node is the less than n
{
//swap n and n1
node temp=*n;
n->age=n1->age;
n->name=n1->name;
n1->age=temp.age;
n1->name=temp.name;
}
}
}
}