Merge Sort for Linked Lists

QUESTION

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.\n\nLet 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.

“TESTCASE_1”: “5\n5 4 4 3 2\n###—###SEPERATOR—###—\n2 3 4 4 5”, “TESTCASE_2”: “6 \n1 3 2 5 6 9\n###—###SEPERATOR—###—\n1 2 3 5 6 9”, “TESTCASE_3”: “4\n2 1 4 3\n###—###SEPERATOR—###—\n1 2 3 4”, “TESTCASE_4”: “7\n1 2 6 3 9 4 5\n###—###SEPERATOR—###—\n1 2 3 4 5 6 9”, “TESTCASE_5”: “0\n###—###SEPERATOR—###—\n0

ANSWER

// C code for linked list merged sort
#include<stdio.h>
#include<stdlib.h>
 
/* 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 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 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;
 int n,i;
  scanf("%d",&n);
  int arr[n];
  for(i=0;i<n;i++)
    scanf("%d",&arr[i]);
  for(i=n-1;i>=0;i--)
    push(&a,arr[i]);

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

printList(a);         
 

return 0;
}
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