Sorting a Singly Linked List With Pointers

问题

I am trying to sort a singly linked list using bubble sort by manipulating ONLY the pointers, no keys.

The following gets stuck in the for loop and loops infinitely. I don't understand why this is. Can anybody explain to me why the end of the list is not being found?

Node* sort_list(Node* head)
{
    Node * temp;
    Node * curr;
    for(bool didSwap = true; didSwap; ) {
            didSwap = false;
            for(curr = head; curr->next != NULL; curr = curr->next) {
                    if(curr->key > curr->next->key) {
                            temp = curr;
                            curr = curr->next;
                            curr->next = temp;
                            didSwap = true;
                    }
                    cout << curr->next->key << endl;
            }
    }
    return head;

}

If I change the code so that the keys (data) are swapped, then the function works properly but for some reason I am not able make it work by manipulating only pointers.

回答1:

Logical Error, you are creating an infinite loop with following code -

temp = curr;
curr = curr->next;
curr->next = temp;

I,e next_of_current is pointing to current, so curr->next will always be curr and never will be NULL;

Next you should use previous pointer to fix your list because your list can be traversed in a single direction. So, Think - If A->B->C->NULL; and you make C and B swap then the new list will still point to A->B and next iteration will be wrong ... because you are not modifying your previous next.

So, another implementation may be -

Node* sort_list(Node* head) {
    Node * curr;
    Node * prev;
    for(bool didSwap = true; didSwap; ) {
        didSwap = false;
        prev = head;
        for(curr = head; curr->next != NULL; curr = curr->next) {
                if(curr->key > curr->next->key) {
                        if (head == curr) {
                            head = curr->next;      
                            curr->next = head->next; 
                            head->next = curr; 
                            prev = head;
                        } else {
                            prev->next = curr->next;
                            curr->next = prev->next->next;
                            prev->next->next = curr
                        }
                        didSwap = true;
                } else if (head != curr) {
                    prev = prev->next;
                } 
                //cout << curr->next->key << endl; // <- this may cause crash if curr->next now points to NULL; (i,e last element)
            }
    }
    return head;
}

Hope this helps, regards.

回答2:

You have following problem:

Let you have list with three members: ptr1->ptr2->ptr3. Before swap you have following pointer set: curr=ptr1; curr->next=ptr2; curr->next->next=ptr3. When you perform swap you receive curr=ptr2; curr->next=ptr1; curr->next->next=ptr2.

E.g. you lost ptr3. You need to change code of inner loop with following:

temp = curr;
temp->next = curr->next->next;    // Save ptr3
curr = curr->next;
curr->next = temp;
didSwap = true;

回答3:

The field you want to swap is the value. However, if you swap the node, the next field will change, the question becomes a little more complex, you need keep the next field right. In a word, change value is a simple and good method.

回答4:

node *sorted_list(node *head) {
    node *index1,*index2;
    for(index1=head;index1->next!=NULL;index1=index1->next) {
        for(index2=index1->next;index2!=NULL;index2=index2->next) {
            if(index1->data>index2->data) {
                int temp=index1->data;
                index1->data=index2->data;
                index2->data=temp;
            }
        }
    }
    return head;
}

来源：https://stackoverflow.com/questions/19556427/sorting-a-singly-linked-list-with-pointers