Merge Sort On Linked List in C - c

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.
I have been struggling to do Merge Sort on a linked list. It keeps throwing back an error. I'm providing the code I've tried to execute. Please do help me out.
It keeps giving runtime error.
#include <stdio.h>
#include <stdlib.h>
struct node {
int data;
struct node *next;
};
struct node *SortedMerge(struct node *a, struct node *b);
void FrontBackSplit(struct node *source, struct node *frontref, struct node *backref);
struct node *Create(struct node *head, int num) {
struct node *newnode, *temp;
newnode = (struct node *)malloc(sizeof(struct node));
newnode->data = num;
newnode->next = NULL;
if (head == NULL) {
head = newnode;
temp = newnode;
} else {
temp->next = newnode;
temp = temp->next;
}
temp->next = NULL;
return head;
}
struct node *display(struct node *head) {
struct node *temp;
temp = head;
while (temp != NULL) {
printf("%d->", temp->data);
temp = temp->next;
}
printf("NULL");
return head;
}
struct node *MergeSort(struct node *head) {
struct node *headref, *a, *b;
headref = head;
if ((head == NULL) || (head->next) == NULL) {
return;
}
FrontBackSplit(headref, a, b);
MergeSort(a);
MergeSort(b);
head = SortedMerge(a, b);
return head;
}
void FrontBackSplit(struct node *source, struct node *frontref, struct node *backref) {
struct node *fast, *slow;
slow = source;
fast = source->next;
while (fast != NULL) {
fast = fast->next;
if (fast != NULL) {
slow = slow->next;
fast = fast->next;
}
}
frontref = source;
backref = slow->next;
slow->next = NULL;
}
struct node *SortedMerge(struct node *a, struct node *b) {
struct node *result;
result = NULL;
if (a == NULL) {
return (b);
}
else if (b == NULL) {
return (a);
}
if (a->data <= b->data) {
result = a;
result->next = SortedMerge(a->next, b);
} else {
result = b;
result->next = SortedMerge(a, b->next);
}
return result;
}
int main() {
struct node *head = NULL;
int i, n, num;
scanf("%d", &n);
for (i = 0; i < n; i++) {
scanf("%d", &num);
head = Create(head, num);
}
head = MergeSort(head);
display(head);
}

Answer 1

There are a couple of problems with the code, and which one triggers the error you are seeing I cannot say, but I will point out a few of them below. Take Create():
struct node *Create(struct node *head, int num)
{
struct node *newnode, *temp;
newnode=(struct node *)malloc(sizeof(struct node));
newnode->data=num;
newnode->next=NULL;
if(head==NULL) {
head=newnode;
temp=newnode;
} else {
temp->next=newnode;
temp=temp->next;
}
temp->next=NULL;
return head;
}
I cannot work out exactly what it is supposed to do, to be honest. Maybe add a new node to a list, represented by a head link? It doesn't do that. You create a new node
newnode=(struct node *)malloc(sizeof(struct node));
which I would suggest you write as
newnode = malloc(sizeof *newnode);
You don't need to cast void *, so you don't need to cast the result of malloc(), and using sizeof *newnode rather than sizeof(struct node) is safer. But the code works correctly in the form you have, so there is not a problem there. However, what happens with that node depends on head. If head is NULL, you point it at the new node, and through temp you (re)assign the new node's next to NULL. So now you will return an updated head that consists of the new node as a single element list. That matches my guess at what the function should do.
However, if head is not NULL, you put the new node in temp->next, which is a problem, since temp isn't initialised. You write to temp in the if(head==NULL) branch, but you dereference it in the else branch, where it can point anywhere. I am surprised if you don't get a segmentation fault from time to time here. It isn't necessary to assign the new node to temp->next, though, because immediately afterwards you change temp to point to temp->next, which is where you just put newnode, so temp = newnode would do the trick, without the segfault. But not all is well if we do that. We now would have the new node in temp (with the next pointer, again, reassigned to NULL) and then we return head. We didn't connect head with newnode anywhere, if we took the else branch. So calling Create() with a non-NULL head creates a new node, throws it away (and leaking memory), and that is all that does.
So while my guess is that Create() should add a new to a list, represented by head, or something to that effect, what it actually does is create a single-element list if the first argument is NULL, and leak sizeof(struct node) memory while doing nothing if head != NULL.
That being said, the code might work by pure luck of course. When I tried it with clang with zero optimisation, I somehow managed to build a list correctly. This is luck, though. It won't work in general. I suspect that what happens is that the repeated calls to Create() in the loop in main() happens to leave the last node you created (and wrote to temp) at the same stack location as the uninitialised temp in the next call. So by pure luck, putting the new node in temp's next appends the new node to the last node you created. It was really interesting working that one out :) But don't rely on this, of course. It is a combination of several lucky circumstances. Add optimisation flags, and the compiler will change the stack layout, and the code will break. Call other functions between successive calls to Create() and the stack will change, and then you don't have the last link on the stack any longer. And the code will break. It is a very unstable situation if this works at all.
If you just want to add a new node to a list, make a prepend function. Something like
struct node *prepend(int val, struct node *list)
{
struct node *n = malloc(sizeof *n);
if (n) {
n->data = val;
n->next = list;
}
return n;
}
(I haven't tested it, so there might by syntax errors, but it will be something like that...you need to figure out what to do if malloc() fails, but you could just abort() if you don't want to deal with it).
There is nothing wrong with display(), except that I don't understand why it is in lower-case when the other functions are in camel-case. You don't need temp, you can use head in the while-loop, but that is a style choice. The function works as intended.
With MergeSort(), however, we have another problem. I am surprised that your compiler didn't scream warnings at you here. It should really give you an error, with the right flags, but at the very least an error. When you test if the list is empty or a singleton, you return, but not with a node. The function should return a struct node *, so just using return will not give you anything useful.
if((head==NULL) || (head->next)==NULL){
return;
}
If the base case of the recursion returns garbage, obviously the whole recursion tumbles. Otherwise, assuming that the FrontBackSplit() and SortedMerge() work, the function looks okay. You don't need the extra headref variable, it is just a synonym for head, but there is nothing wrong with having it. The compiler will get rid of it for you. There isn't any need to assign the merged lists to head and then return head either. You can just return SortedMerge(a,b). But again, your compiler will handle that for you, once you turn on optimisation. Except for the base case, I believe the function should work.
In FrontBackSplit(), I get the impression that you want to get the frontref and backref values back to the caller. Otherwise, the function doesn't do anything. But when you are modifying the function parameters, you are not changing the variables in the caller's scope. You need to pass the two pointers by reference, which means that you need to use pointers to pointers. Change it to something like this:
void FrontBackSplit(struct node *source,
struct node **frontref,
struct node **backref)
{
struct node *fast, *slow;
slow=source;
fast=source->next;
while(fast!=NULL) {
fast=fast->next;
if(fast!=NULL) {
slow=slow->next;
fast=fast->next;
}
}
*frontref=source;
*backref=slow->next;
slow->next=NULL;
}
When you call the function, use the addresses of the parameters for the second and third argument, so use FrontBackSplit(headref,&a,&b); instead of FrontBackSplit(headref,a,b);.
As far as I can see, SortedMerge() should work (with a modified FrontBackSplit()). It is recursive, but not tail-recursive, so you might have problems with overflowing the stack for long lists. It isn't hard to make iterative, though.
You should make main() either int main(void) or int main(int, char**). You should return 0 for success.
My guess is that one of three things are breaking your code. When you Create() your lists, you do not get the lists you want. In just the right circumstances, with just the right compiler and function call configurations, however, you might get lucky (and maybe that is what you have seen). In that case, it might be the return in MergeSort(). Return head instead, there, that is probably what you want. If you have an empty list or a list of length one, you should return that list. So change return; to return head;. And if it isn't that either, it is probably because you recurse on random data in MergeSort(), because a and b aren't initialised in the recursion. They are uninitialised when you call FrontBackSplit() and the call doesn't change them, because they are passed by value and not reference. The change I listed above will fix that.
There might be more that I have overlooked, but at least those three issues are enough to break the code, each of them on their own, so it is a good place to start with debugging.