Linked List Merge Sort Exaplanation

Question

Can someone explain to me please, how this code works :

http://www.chiark.greenend.org.uk/~sgtatham/algorithms/listsort.c

I don't understand the algorithm used in this post. Thanks

Answer 1

Merge sort is often preferred for sorting a linked list. The slow random-access performance of a linked list makes some other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible.

Let head be the first node of the linked list to be sorted and headRef be the pointer to head. Note that we need a reference to head in MergeSort() as the below implementation changes next links to sort the linked lists (not data at the nodes), so head node has to be changed if the data at original head is not the smallest value in linked list.

MergeSort(headRef)

1) If head is NULL or there is only one element in the Linked List then return.

2) Else divide the linked list into two halves. FrontBackSplit(head, &a, &b); /* a and b are two halves */

3) Sort the two halves a and b. MergeSort(a); MergeSort(b);

4) Merge the sorted a and b (using SortedMerge() discussed here) and update the head pointer using headRef. *headRef = SortedMerge(a, b);

    /* Link list node */
    struct node
    {
        int data;
        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->data 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 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 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)
      {
       printf("%d ", node->data);
       node = node->next;
      }
    }

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

    /* Drier program to test above functions*/
    int main()
    {
      /* Start with the empty list */
      struct node* res = NULL;
      struct node* a = NULL;
      struct node* b = 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);
      push(&a, 10);
      push(&a, 5);
      push(&a, 20);
      push(&a, 3);
      push(&a, 2); 

      /* Remove duplicates from linked list */
      MergeSort(&a);

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

      getchar();
      return 0;
    }

Answer 2

Try imaging all the merges that are performed in a normal merge sort on an array: first, elements are paired up and merged into sorted subarray of length two, then these subarray of length two are paired up and merged into sorted subarray of length four and so on. Notice the length of the subarray: 1, 2, 4, and so on, let's call this instep, which doubles in each iteration.

At any point, p points to a list of length instep, q points to a list of length instep or smaller (we may hit the end of the list), and q immediately follows p. They form a pair of subarray as mentioned above. We do a merge on p and q to get a sorted list of length psize + qsize starting from p. We than move p and q to the next pair, and so on. Once we are done with the whole list, we double instep and start merging longer sorted list.