C Program for Merge Sort for Linked Lists

17 Dec, 2021
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 the 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 the head node has to be changed if the data at the original head is not the smallest value in the 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);
C
// C code for linked list merged sort#include <stdio.h>#include <stdlib.h> // Link list nodestruct Node{    int data;    struct Node* next;}; // Function prototypesstruct 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 https:// www.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 <= 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 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;    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)    {        printf("%d ", node->data);        node = node->next;    }} /* Function to insert a node at the   beginning 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;} // Driver codeint 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);    push(&a, 10);    push(&a, 5);    push(&a, 20);    push(&a, 3);    push(&a, 2);     // Sort the above created Linked List    MergeSort(&a);     printf("Sorted Linked List is: \n");    printList(a);     getchar();    return 0;}
Output:
Sorted Linked List is: 
2 3 5 10 15 20
Time Complexity: O(n*log n)
Space Complexity: O(n*log n)
Please refer complete article on Merge Sort for Linked Lists for more details! 
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