Reputation: 165
I'm getting mad with infinite loop, what do you think is suitable solution?
void sorting () {
node * temphead = head;
node * tempnode = NULL;
for (int i=0; i<count; i++) {
for (int j=0; j<count-i; j++) {
if (temphead->data > temphead->next->data) {
tempnode = temphead;
temphead = temphead->next;
temphead->next = tempnode;
}
temphead=temphead->next;
count++;
}
}
}
I tried to increment count and use many conditions with while- before and after the for loop with no result
Upvotes: 2
Views: 11509
Reputation: 11439
An easier way to slide through a linked list is like this:
for (node *current = head; current != nullptr; current = current->next) {
// This will run through all of the nodes until we reach the end.
}
And to slide to the second to last item (ensuring that node->next
exists) looks like this:
for (node *current = head; current->next != nullptr; current = current->next) {
// Go through all of the nodes that have a 'next' node.
}
If you want to count how many items are in a linked list, you do something like this:
int count = 0;
for (node *current = head; current != nullptr; current = current->next) {
count = count + 1;
}
So a selection type sort like you have above would look like this:
for (node *index = head; index->next != nullptr; index = index->next) {
for (node *selection = index->next; selection != nullptr; selection = selection->next) {
if (index->data > selection->data) {
swap(index->data, selection->data);
}
}
}
Although sorting linked lists is generally not the best way to go (unless you're performing a merge).
Upvotes: 5
Reputation: 205
the problem is you are looping till count and you are incrementing count in every run of the loop//remove line count++ to avoid remove infinite loop
Upvotes: 2