在链表中对偶数和奇数进行排序

Question

Im trying to write a function that takes a random list and sorts it (in c) so that even numbers are first, doing so without creating a second list. What im trying to do is moving a newly found even number to the head of the list (making the previous head the seconnd element). Cant understand why my pointers get messed up though -

typedef struct node node;
typedef struct node {
    int data;
    node *next;
};

void sortlist(node *head) {
    if (head == NULL) {
        printf("Empty list \n");
    }
    node *index1 = head->next;
    node *oldhead;
    if (index1 == NULL) {
        return;
    }
    while (index1 != NULL) {
        if (index1->data % 2 == 0) {
            oldhead = head;
            head = index1;
            head->next = oldhead;
        }
        index1 = index1->next;
    }
}

Answer 1

Main idea : Suppose you have a list as below.

1->3->2->4

When you want to make the element 2 the new head, you have to do two things.

• Unlink element 2 and make next element of 3 to 4 . [which you are missing in your code]
• Make a new node and make it the new head of the list.

Then the list will look like 2->1->3->4 . Again doing the same thing for element 4 , it will look like 4->2->1->3 .

So, I believe you need (tentatively) something like this.

void sortlist(node *head) {
    if (head == NULL) {
        printf("Empty list \n");
    }
    node *current = head->next;
    node *previous = head;
    while(current != NULL){
        if(current->data % 2 == 0){
            previous->next = current->next;
            current->next = head;
            head = current;
        } else{
            previous = previous->next;
            //previous = current; [an equivalent statement]
        }
        current = previous->next;
    }
}

An alternative approach if you want to create a new node and make it the head of the list.

void sortlist(node *head) {
    if (head == NULL) {
        printf("Empty list \n");
    }
    node *current = head->next;
    node *previous = head;
    while(current != NULL){
        if(current->data % 2 == 0){
            previous->next = current->next;
            node* new_node = (struct node*) malloc (sizeof (struct node));
            new_node->data = current->data;
            new_node->next = head;
            head = new_node;
        } else{
            previous = previous->next;
        }
        current = current->next;
    }
}

Update : I have verified both the code snippet. Its working!

Answer 2

This is not enough: when you move a node to the head, you need also to direct its previous to its next . Since your list is linked in only one direction, you need to keep track of what was its previous.

Moreover, sending the head pointer like this, by value , will not change it in the caller, so the caller will loose track of the head of the list. Therefore, you need to pass the head pointer in a way that it can be modified and retrieved, that is, send a pointer to that pointer:

void sortlist(node** head) {
    if (*head == NULL) {
        printf("Empty list \n");
    }
    node *index1 = (*head)->next;
    node *prev = *head;

    while (index1 != NULL) {
        if (index1->data % 2 == 0) {
            prev->next = index1->next;
            index1->next = *head;
            *head = index1;
        }
        else {
            prev = index1;
        }
        index1 = prev->next;
    }
}

Answer 3

It is not clear why you skip the head itself and start with head->next.

Also the head node should be passed by reference because the head is changed in the function.

And it does not make sense to output a message that the list empty because the caller can check itself before calling the function whether the list is empty can not it?

The function can look the following way

void sortlist(node **head)
{
    for (node **current = head; *current != NULL; )
    {
        if ((*current)->data % 2 == 0 && current != head)
        {
            node *tmp = *current;
            *current = (*current)->next;
            tmp->next = *head;
            *head = tmp;
        }
        else
        {
            current = &(*current)->next;
        }
    }
}

For example if the list contains the following clause

0 1 2 3 4 5 6 7 8 9

then after calling the function it will look like

8 6 4 2 0 1 3 5 7 9