Selection sort with linked list's - c

Question

I have the following data structure:
struct scoreentry_node {
struct scoreentry_node *next;
int score;
char name[1];
};
typedef struct scoreentry_node *score_entry;
I am trying to create a function that consumes my structure in order and arranges them in ascending order based on the name. I want to modify the input without allocating any memory or freeing anything:
I've tried your suggestions:
void selectionsort(score_entry *a) {
for (; *a != NULL; *a = (*a)->next) {
score_entry *minafteri = a;
// find position of minimal element
for (score_entry j = (*a)->next; j != NULL; j = j->next) {
if (strcmp(j->name, (*minafteri)->name) == -1) {
*minafteri = j;
}
}
// swap minimal element to front
score_entry tmp = *a;
a = minafteri;
*minafteri = tmp;
}
}
I'm testing the above code with the following:
score_entry x = add(8, "bob", (add( 8 , "jill", (add (2, "alfred", NULL)))));
iprint("",x);
selectionsort(&x);
iprint("", x);
clear(x); //Frees the whole list
iprint() prints the score and name fields in the struct. My add function is as follows:
score_entry add(int in, char *n, score_entry en) {
score_entry r = malloc(sizeof(struct scoreentry_node) + strlen(n));
r->score = in;
strcpy(r->name, n);
r->next = en;
return r;
}
I'm getting heap errors and my second print doesn't print the sorted list, it prints nothing. What am I doing wrong, and what can I do to fix it?

Answer 1

besides passing the pointer by address (see comments below) you also need to fix the way you swap elements too
void selectionsort(score_entry *a) {
for (; *a != NULL; *a = (*a)->next)
{
score_entry *minafteri = a;
// find position of minimal element
for (score_entry j = (*a)->next; j != NULL; j = j->next) {
if (strcmp(j->name, (*minafteri)->name) == -1) {
*minafteri = j;
}
}
// swap minimal element to front
score_entry tmp = *a;
a = minafteri; // put the minimal node to current position
tmp->next = (*a)->next ; //fix the links
(*minafteri)->next=tmp; //fix the links
}
}

Answer 2

You have to pass the argument to selectionsort by reference:
void selectionsort(score_entry *a) {
for (; *a != NULL; *a = (*a)->next)
{
score_entry *minafteri = a;
// find position of minimal element
for (score_entry j = (*a)->next; j != NULL; j = j->next) {
if (strcmp(j->name, (*minafteri)->name) == -1) {
*minafteri = j;
}
}
// swap minimal element to front
score_entry tmp = *a;
a = minafteri;
*minafteri = tmp;
}
}

Answer 3

This code is hideous! Not only have you not provided us with all of the necessities to reproduce your problem (we can't compile this!), but you've hidden pointer abstractions behind typedef (also a nightmare for us). Generally speaking, one shouldn't even use linked lists in C anymore let alone sort them...
Nonetheless, there are two answers here.
*minafteri = j; found within your find loop actually modifies your list! Why should your find loop be modifying your list?
Answer: it shouldn't! By instead assigning minafteri = &j->next, you won't be modifying the list with your find loop...
Alternatively, you could perform the swap inside of that loop.
*minafteri = j; would need to swap the following, in this order:
(*minafteri)->next and j->next
*minafteri and j
Do you think that single line of code is capable of performing those two swaps? Well, it gets half way through one of them... and removes a heap of elements from your list in the process!
The following appears to be a faulty attempt swapping elements:
score_entry *minafteri = a; // half of assigning `a` to `a`
/* SNIP!
* Nothing assigns to `minafteri` in this snippet.
* To assign to `minafteri` write something like `minafteri = fubar;` */
score_entry tmp = *a; // half of assigning `*a` to `*a`
a = minafteri; // rest of assigning `a` to `a`
*minafteri = tmp; // rest of assigning `*a` to `*a`
It's really just assigning *a to *a and a to a... Do you think you need to do that?
I'd have thought you'd notice that when you were creating your MCVE... Ohh, wait a minute! Shame on you!
Focus on swapping two nodes within a list as a smaller task. Once you've done that, consider taking on this task.

Answer 4

There are multiple problems in your code:
if (strcmp(j->name, (*minafteri)->name) == -1) { is incorrect: strcmp() does not necessarily return -1 when the first string is less than the second, it can return any negative value.
The way you adjust the links in order to move the lower entry is incorrect: you cannot update the link from the previous node to the one you move to the start. The list is corrupted.
Here is an improved version:
void selectionsort(score_entry *a) {
for (; *a != NULL; a = &(*a)->next) {
// find position of minimal element
score_entry *least = a;
for (score_entry *b = &(*a)->next; *b != NULL; b = &(*b)->next) {
if (strcmp((*b)->name, (*least)->name) < 0) {
least = b;
}
}
if (least != a) {
// swap minimal element to front
score_entry n = *least;
*least = n->next; /* unlink node */
n->next = *a; /* insert node at start */
*a = n;
}
}
}

Answer 5

Here is the Java Implementation of Selection Sort on Linked List:
Time Complexity: O(n^2)
Space Complexity: O(1) - Selection sort is In-Place sorting algorithm
class Solution
{
public ListNode selectionSortList(ListNode head)
{
if(head != null)
{
swap(head, findMinimumNode(head));
selectionSortList(head.next);
}
return head;
}
private void swap(ListNode x, ListNode y)
{
if(x != y)
{
int temp = x.val;
x.val = y.val;
y.val = temp;
}
}
private ListNode findMinimumNode(ListNode head)
{
if(head.next == null)
return head;
ListNode minimumNode = head;
for(ListNode current = head.next; current != null; current = current.next)
{
if(minimumNode.val > current.val)
minimumNode = current;
}
return minimumNode;
}
}