Delete all elements with same value from doubly circular linked list

Question

I have doubly circular linked list

struct Element {
    int data;
    Element *next;
    Element *previous;
};

struct List {
    Element *head;
};

The task is to delete elements/nodes with the same data. For example, if the list has 1,1,3,4,5,1 and after parsing the function value=1 it should be as following 3,4,5.

bool removeAllValue(List &l, int value) {
    if(!l.head)
        return false;
    else if(l.head->next == l.head && l.head->data == value){
        l.head = NULL;
        return true;
    }
    Element *p = l.head;
    do{
        if(p->data == value)
            removeElement(l, p);
        p = p->next;
    }while(p!=l.head);
    return true;

Above is function which itterates through elements, finds element with given value/data and calls the following function to delete element.

void removeElement(List &l, Element *element){
    Element *previous = element->previous;
    Element *next = element->next;
    Element *head = l.head;
    if(next == head){ //if element is tail of the list
        l.head->previous = element->previous; // relinking head's previous to new tail
        element->previous->next = head;
    }else if(element == head){ // if element is head of the list
        l.head = next; // relinking head to tail
        l.head->previous = previous; // relinking tail to head
    }else{
        previous->next = next;
        next->previous = previous;
    }
    delete element;

}

I have tried different ways but it always gave me errors. The problem is in relinking tail and head, as I think but I can't find the way to fix it.

Answer 1

The problem was in few relinking operations

void removeElement(List &l, Element *element){
    Element *previous = element->previous;
    Element *next = element->next;
    Element *head = l.head;
    if(element->next == element)
        l.head = NULL;
    else if(next == head){ //if element is tail of the list
        Element *newTail = previous;
        newTail->next = l.head;
        l.head->previous = newTail;
    }else if(element == head){ // if element is head of the list
        l.head = next; // relinking head
        l.head->previous = previous; // relinking tail to head
        previous->next = l.head; // relinking head to tail
    }else{
        previous->next = next;
        next->previous = previous;
    }
    delete element;
}

And as Marek Fekete noticed I didn't change iterating pointer each time I deleted an element. So I just iterated through all list from the beginning, each time I deleted an element.

bool removeAllValue(List &l, int value) {
    if(!l.head)
        return false;
    else if(l.head->next == l.head && l.head->data == value){
        l.head = NULL;
        return true;
    }
    Element *p = l.head;
    bool end = false;
    while(!end){
        do{
            if(p->data == value){
                Element *tmp = p;
                p = p->next;
                removeElement(l, tmp);
                break;
            }else
                p = p->next;
            if(p==l.head || l.head == NULL){
                end = true;
                break;
            }
        }while(p);
    }
    return true;
}

Answer 2

I think you can streamline it. Handling the removing tail/head/middle situations in removeElement differently is not needed, with the only extra of resetting the list's head when removing the former head. I would rather put determining the next element in list and last item removal detection. Like this:

Element* removeElement(List &l, Element *element) {
    Element *newnext;
    Element *previous = element->previous;
    Element *next = element->next;
    Element *head = l.head;

    previous->next = next;
    next->previous = previous;
    newnext = next;

    if (element == head) { // if element is head of the list
        l.head = next; // set the head correctly
    }

    if (element == newnext)  // detect if last element in the list was deleted
        l.head = newnext = nullptr;
    else if (newnext == head) // detect if next is at the head again
        newnext = nullptr;

    delete element;
    return newnext;
}

Then you do not have to re-iterate from the head each tiem you remove an element and you do not have to handle the single item list as a special case:

bool removeAllValue(List &l, int value) {
    if (!l.head)
        return false;

    Element *p = l.head;
    do {
        if (p->data == value)
        {
            p = removeElement(l, p);
        }
        else
        {
            p = p->next;
            if (p == l.head)
                p = nullptr;
        }
    } while (p);
    return true;
}