Question

(Write a C# program DO NOT USE CLASS)Implement the merge sort algorithm using a linked list instead of arrays. You can use any kind of a linked structure, such as single, double, circular lists, stacks and/or queues. You can populate your list from an explicitly defined array in your program. HINT: You will not be using low, middle and high anymore. For finding the middle point, traverse through the linked list while keeping count of the number of nodes. Break up the list into two, null terminated lists, based on the node count. //Here is a portion of what mergeSort function would look like: //a points to the left partition, b points to the right partition. They would be passed by reference. //You would then recursively process each partition. if(head==NULL) return NULL; if (head->next==NULL) return head; a=mergeSort(a); b=mergeSort(b); c=merge(a,b); return (c); //These are the function headers void split (node* head,node*&a,node*&b) node* merge(node* a, node* b) node* mergeSort( node* head) Make sure to account for when head is null or when there is only one item in the list.

Answer 1

// linked list merged sort by C++
using namespace std;

/* Link list node */
class Node {
public:
   int data;
   Node* next;
};

/* function prototypes */
Node* SortedMerge(Node* a, Node* b);
void FrontBackSplit(Node* source,
                   Node** frontRef, Node** backRef);

/* sorts the linked list by changing next pointers (not data) */
void MergeSort(Node** headRef)
{
   Node* head = *headRef;
   Node* a;
   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);
}
Node* SortedMerge(Node* a, Node* b)
{
   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->data <= b->data) {
       result = a;
       result->next = SortedMerge(a->next, b);
   }
   else {
       result = b;
       result->next = SortedMerge(a, b->next);
   }
   return (result);
}

/* UTILITY FUNCTIONS */
/* Split the nodes into front and back halves,
   and return two lists using the reference parameters.
   If the length is odd, the extra node should go in the front list.
we can Use the fast/slow pointer strategy. */
void FrontBackSplit(Node* source,
                   Node** frontRef, Node** backRef)
{
   Node* fast;
   Node* slow;
   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(Node* node)
{
   while (node != NULL) {
       cout << node->data << " ";
       node = node->next;
   }
}

/* Function to insert a node at the beginging of the linked list */
void push(Node** head_ref, int new_data)
{
   /* allocate node */
   Node* new_node = new Node();

   /* put in the data */
   new_node->data = new_data;

   /* 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;
}

/* Driver program to test above functions*/
int main()
{
   /* Start with the empty list */
   Node* res = NULL;
   Node* a = NULL;

   /* unsorted linked lists to test the functions
Created lists shall be a: 2->3->20->5->10->15 */
   push(&a, 15);
   push(&a, 10);
   push(&a, 5);
   push(&a, 20);
   push(&a, 3);
   push(&a, 2);

   /* Sort the above created Linked List */
   MergeSort(&a);

   cout << "Sorted Linked List is: \n";
   printList(a);

   return 0;
}

OUTPUT :

Sorted Linked List is: 
2 3 5 10 15 20