Question

我是新手，因此我几乎没有学习如何实现链接列表。当我输入数字时，我的程序崩溃。我想将节点添加到列表的后面。

这是节点

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

这是添加功能

void Add(node* &head, int newdata)
{
    //create a new node to hold the data with a terminal (NULL) next pointer           
       node *tmp = new node;
       tmp->data = newdata;
       tmp->next;
       node *current = head;
    //check whether head has been initialized (is NULL)
    // if not, make the new node head and set prev
       if ( head != NULL)
       {
            tmp = head;
            tmp->prev = NULL;
       } 

    //if head has been initialized
    //find the end of the chain with a pointer
      else
      {
           while (current->next != NULL )
           {
                current = current->next;
           }
      }

    //add the new node on to the last node in the list
    //set pointers both forward and backwards
     tmp = current;
     tmp->prev = current->prev->next;
     tmp->next = current->next;



}

Answer 1

开始时，您需要将tmp->next和tmp->prev设置为NULL。垃圾指针每次都会杀了你。

然后你似乎认为head == NULL意味着它已被初始化，而它可能意味着相反。

最后，您设置了tmp = current;，因此您将丢弃要添加的节点。

尝试再次完成最后三行。

另外，请使用调试器进行操作，以防您无法查看自己正在做的事情。

Answer 2

嗯，你在add函数中有一些奇怪的东西，这可能有助于你理解这个过程：

 void add(node* &head, int dataToAdd){
    node* newNode = new node();
    newNode->data = dataToAdd;

    if(!head){ // checking to see if head was passed in or is null
        head = newNode;
        return;
    }

    node* current = head;
    node* next = current->next;
     // iterate through the list till next == Null( hits 1 after the end)
    while(current->next){
       current = next;
       next = next->next;
    }

    //Set the end of the list to the newly added node
    next = newNode;
    next->prev = current;

 }

Answer 3

这条线应该做什么？

tmp->next;

使用链接列表可能会不时进行真正的大脑练习。我建议你直接将指针初始化为NULL。关于最后三行，他们需要再次考虑。

 tmp = current; //This discards your newly created node, which results in a memory leak
 tmp->prev = current->prev->next; // The previous node is simply 'current'
 tmp->next = current->next; // You know that the next node will be NULL

current节点还需要知道新的current->next将是什么。

Answer 4

你需要做更多这样的事情：

void Add(node* &head, int newdata)
{
    //create a new node to hold the data with a terminal (NULL) next pointer           
    node *tmp = new node;
    tmp->data = newdata;
    tmp->next = NULL;
    tmp->prev = NULL;

    if (!head)
        head = tmp;
    else
    {
        //find the end of the chain with a pointer
        node *last = head;
        while (last->next != NULL )
            last = last->next;

        //add the new node on to the last node in the list
        //set pointers both forward and backwards
        last->next = tmp;
        tmp->prev = last;
    }
}

如果你跟踪最后一个节点以及头部，你的插入将会快得多，因为你不必每次迭代整个列表，例如：

struct list
{
    node *head;
    node *tail;
};

void Add(list &l, int newdata)
{
    node *tmp = new node;
    tmp->data = newdata;
    tmp->next = NULL;

    if (!l.head)
        l.head = tmp;

    tmp->prev = l.tail;
    if (tmp->prev)
        tmp->prev->next = tmp;

    l.tail = tmp;
}

话虽如此，因为您使用的是C ++，所以您真的应该使用std::list，它会为您处理所有这些：

#include <list>

std::list<int> mylist;
mylist.push_back(12345);
// etc...

c ++添加到双向链表的末尾

4 个答案: