C ++删除类似节点的链表

Question

For a homework assignment I need to remove all similar nodes that the number passed into. For example if I have on the list

3 5 5 4

the 5's will be removed from the linked list and I will end with

3 4

we are not allowed to use the std library for this class and here is the header file

    namespace list_1
{
    class list
    {
    public:
        // CONSTRUCTOR
        list( );
        // postcondition: all nodes in the list are destroyed.
        ~list();
        // MODIFICATION MEMBER FUNCTIONS
        //postcondition: entry is added to the front of the list
        void insert_front(const int& entry);
        //postcondition: entry is added to the back of the list
        void add_back(const int& entry);
        // postcondition: all nodes with data == entry are removed from the list
        void remove_all(const int& entry);
        // postcondition: an iterator is created pointing to the head of the list
        Iterator begin(void);

        // CONSTANT MEMBER FUNCTIONS
        // postcondition: the size of the list is returned
        int size( ) const;
    private:
        Node* head;
    };

}

I can understand how to remove the front, and the back of the list. But for some reason I can't wrap my head around going through the list and removing all of the number that is passed in. Anything helps! Thanks

edited to include Node.h

#pragma once

namespace list_1
{
    struct Node
    {
        int data;
        Node *next;

        // Constructor
        // Postcondition: 
        Node (int d);
    };
}

Answer 1

There are two ways of doing this. The first is to iterate through the list and remove the nodes. This is tricky because to do that you have to keep a pointer to the previous node so you can change its next value. The code for removing a node would look like this (assume current is the current node and prev is the previous node)

Node* next = current->next;
delete current;
prev->next = next;

Maintaining a reference to the previous node can be a bit tedious though, so here is another way to do it. In this method, you essentially create a new list but don't insert Nodes who's data is equal to entry .

The code might look a little like this

void list::remove_all(const int &entry)
{
    Node* newHead = NULL;
    Node* newTail = NULL;
    Node* current = head;

    // I'm assuming you end your list with NULL
    while(current != NULL)
    {
        // save the next node in case we have to change current->next
        Node* next = current->next;
        if (current->data == entry)
        {
            delete current;
        }
        else
        {
            // if there is no head, the set this node as the head
            if (newHead == NULL)
            {
                newHead = current;
                newTail = current;
                newTail->next = NULL; // this is why we saved next
            }
            else
            {
                // append current and update the tail
                newTail->next = current;
                newTail = current;
                newTail->next = NULL; // also why we saved next
            }
        }
        current = next; // move to the next node
    }
    head = newHead; // set head to the new head
}

Note: I didn't test this, I just typed it up off the top of my head. Make sure it works. =)

Hope this helps! ;)