1

对于家庭作业，我需要删除该数字传入的所有类似节点。例如，如果我在列表中

3 5 5 4

5's 将从链表中删除，我将以

3 4

我们不允许使用这个类的标准库，这是头文件

    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;
    };

}

我可以理解如何删除列表的前面和后面。但是由于某种原因，我无法绕过列表并删除所有传入的数字。任何有帮助！谢谢

编辑以包含 Node.h

#pragma once

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

        // Constructor
        // Postcondition: 
        Node (int d);
    };
}
4

1 回答 1

2

有两种方法可以做到这一点。第一个是遍历列表并删除节点。这很棘手，因为要做到这一点，您必须保留指向前一个节点的指针，以便您可以更改其next值。删除节点的代码如下所示（假设current是当前节点并且prev是前一个节点）

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

维护对前一个节点的引用可能有点乏味，所以这里有另一种方法。在此方法中，您实际上创建了一个新列表，但不插入data等于的节点entry。

代码可能看起来有点像这样

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
}

注意：我没有对此进行测试，我只是从头顶输入了它。确保它有效。=)

希望这可以帮助！;)

于 2013-09-11T00:03:10.437 回答