2

所以我一直在搜索论坛，但我对语言和链接列表仍然很陌生，所以我几乎无法破译结果。

基本上我为我的链表做了一个删除功能。我目前可以创建列表，遍历列表，排序列表，搜索列表，并在链表中的任何节点之前插入。我从插入中回收了一些代码，以找到列表中可以删除的点。我的主要困惑是如何将以前的点链接到我要删除的节点之后的节点。

4

5 回答 5

7

我不会编写一个全新的链表实现，但我可以为您指出代码中的一些问题。

诀窍是在要删除的节点之前保持一个节点。

为了清楚起见，我已重命名entry为current

nodetype *current , *first, *next;
int akey;

// With this line your search will start from the second element.
// current =start->ptr;
// Should be
current = start;

// this is not needed. I am assuming the last node has NULL value for '->ptr'
// last=start;
next = current->ptr;

cout<<"Input Data You Would Like To Delete"<<endl;
cin>>akey;

// Check if the first node contains the data
// Assuming the list will have at least one element. i.e. current is not NULL 
while (current->adata == akey)
{
  // Delete it.
  delete current;
  // Update current for the while loop
  current = next;
  // update next too.
  next = current->ptr;
}
// Now we know the first element doesn't contain the data.
// Update the pointer to the begging of the new list if anything is removed from the top.
first = current;

// This has unnecessary checks. 
// ****Specifically (akey!=current->adata) will 
// prevent you from entering the loop if it is false.
// while((akey!=current->adata)&&(current->ptr !=NULL))
while(next != NULL)  // This should be enough
{
     if(next->adata == akey)
     {
          // make the current node (before the 'deletion point') 
          // lined to the one after the 'deletion point (next of the next)
          current->ptr = next->ptr; 
          // delete the node. 
          delete next;
          // Make the next pointer point to the new next. 
          next = current->ptr
     }
    // Otherwise advance both current and next.
     else {
           current = next;
           next = next->ptr; 
     }
}

// Use this to test it.
current = first;
while(current){
    cout<<current->adata<<", ";
    current = current->ptr;
}

这不是最干净的方法。但是，它与您的实现类似，因此您可以看到哪里出错了。

于 2013-03-24T20:24:04.073 回答
4 
#include <iostream>
#include <string>
                                         // blank line(s) after includes

using namespace std;                     // some people will say to avoid this
                                         // but I use it in examples for brevity

                                         // blank line(s) around class def
class nodetype
{                                        // bracket on its own line
public:                                  // non indented visibility specifier
    nodetype(int value, nodetype *p)     // constructor first declared in class
    {
        adata = value;                   // level of indentation for fn body
        ptr   = p;                       // spaces around operators like =
    }
                                         // blank line(s) between fns and vars
    int adata;
    nodetype *ptr;
};
                                         // blank line(s) between class and fn
void LinkedListDelete(nodetype **start, int akey)
{
    nodetype *current, **previous;       // pointer *s are connected to vars
                                         // blank line between section
    previous = start;
    current  = *start;
                                         // blank line between section
                                         // I use blank lines a lot, they help
                                         // me to organize my thoughts

    while((current != NULL) && (akey != current->adata))
    {                                    // indentation inside nested scope
        previous = &current->ptr;        // no space for unary operators like &
        current  = current->ptr;         // assignments justified to same level
    }

    if (current != NULL)
    {
        *previous = current->ptr;        // no space for unary *, space for =
        delete current;
    }
                                         // more blank lines between sections
    return;
}

void LinkedListPrint(nodetype *list)     // no space for unary *
{                                        // brackets on their own lines
    while (list != NULL)                 // space around !=
    {
        cout << "(Node: " << list->adata << ") ";
        list = list->ptr;                // spaces around <<
    }
    cout << endl;
}

int main()
{
    nodetype *node = new nodetype(5, new nodetype(10, // justified stuff
                     new nodetype(7, new nodetype(14,
                     new nodetype(23, NULL)))));
                                         // blank lines
    cout << "Build linked list: ";
    LinkedListPrint(node);
    cout << "Removed node 7: ";
    LinkedListDelete(&node, 7);
    LinkedListPrint(node);

    return 0;
}

我根据您提供的代码制作了此代码。不完全一样，我改变了一些东西，但它会做你想要的。我不得不猜测 nodetype 的结构是什么，为了方便我添加了一个构造函数。我添加了一些评论，指出我的风格的各个方面。

请注意，它比您最初提供的代码更容易阅读。风格很重要。人们会告诉你必须使用 X 或 Y 风格，但真正重要的是你选择你喜欢的任何风格并始终如一地坚持下去；它将使您更容易快速阅读和理解自己的代码。

相信我，当您编写了大量代码时，您将无法一次记住所有代码，并且能够快速弄清楚您在做什么是必不可少的。

于 2013-03-24T20:18:22.410 回答
0

考虑下面给出的结构，

struct info
{
 int data;
 struct info *next;
};

如果您使用上述结构将记录存储在链表中，则可以使用以下代码从链表中删除元素，

void delitem()
{
  info *curr,*prev;
  int tdata;
  if(head==NULL)
  {
    cout<<"\nNo Records to Delete!!!";
  }
  cout<<"\nEnter the Data to be deleted: ";
  cin>>tdata;
  prev=curr=head;
  while((curr!=NULL)&&(curr->data!=tdata))
  {
    prev=curr;
    curr=curr->next;
  }
  if(curr==NULL)
  {
    cout<<"\nRecord not Found!!!";
    return;
  }
  if(curr==head)
  {
    head=head->next;
    cout<<"\nData deleted: "<<tdata;
  }
  else
  {
    prev->next=curr->next;
    if(curr->next==NULL)
    {
      temp=prev;
    }
    cout<<"\nData deleted: "<<tdata;
  }
  delete(curr);
}
于 2013-03-24T19:48:46.113 回答
0

我认为删除一个节点或在链表中插入 ine 过于简单易行，但需要准确了解其机制。这个例子展示了如何添加和删除节点，但它不是一个完整的程序，但它揭示了添加和删除以及移动链表的机制：

#include <iostream>
using namespace std;

//class Data to store ages. in a real program this class can be any other
//class for example a student class or customer...
class Data
         {
         public:
               Data(int age):itsAge(age){}
               ~Data(){}
               void SetAge(int age){itsAge=age;}
               int  getAge()const{return itsAge;}
         private:
                int itsAge;
         };

//the most important part of the program is the linked0list
class Node
          {
          public:
                //we just make itsPtrHead when created points to Null even if we pass a pointer to Data that is not NULL
                Node(Data* pData): itsPtrData(pData),itsPtrHead(NULL),itsCount(0){}

                Data* getPdata()const;
                Node* getPnext()const;
                int   getCount()const{return itsCount;}

                void  insertNode(Data*);//import bcause it shoes the mechanism of linked-list
                void  deleteNode(int);//most significant in this program
                Data* findData(int&,int);
                void  print()const;
         private:
                Data* itsPtrData;
                Node* itsPtrHead;
                int   itsCount;

          };

Data* Node::getPdata()const
{
if(itsPtrData)
    return itsPtrData;
else
    return NULL;
}
Node* Node::getPnext()const
{
return itsPtrHead;
}
void Node::insertNode(Data* pData)
{
Node* pNode=new Node(pData);//create a node
Node* pCurrent=itsPtrHead;//current node which points first to the first node that is the "head bode"
Node* pNext=NULL;//the next node

int NewAge=pData->getAge();//the new age that is past to insertNode function
int NextAge=0;//next age
itsCount++;//incrementing the number of nodes

//first we check wether the head node "itsPtrHead" points NULL or not
//so if it is null then we assign it the new node "pNode" and return from insert function
if(!itsPtrHead)
    {
    itsPtrHead=pNode;
    return;
    }
//if the condition above fails (head is not null) then check its value and
//compare it with new age and if the new one is smaller than the head
//make this new one the head and then the original node that head points to it make it its next node to the head
if(itsPtrHead->getPdata()->getAge() > NewAge)
    {
    pNode->itsPtrHead=itsPtrHead;
    itsPtrHead=pNode;
    //exit the function
    return;
    }
//if the condition above fails than we move to the next node and so on
for(;;)
    {
    //if the next node to the current node is null the we set it with this new node(pNode) and exit
    if(!pCurrent->itsPtrHead)
        {
        pCurrent->itsPtrHead=pNode;
        return;
        }
    //else if it not null(next node to current node)
    //then we compare the next node and new node
    pNext=pCurrent->itsPtrHead;
    NextAge=pNext->getPdata()->getAge();
    //if next node's age is greater than new then we want new node 
    //to be the next node to current node then make the original next to current to be the next to its next
    if(NextAge > NewAge)
        {
        pCurrent->itsPtrHead=pNode;
        pNode->itsPtrHead=pNext;
        //exitting
        return;
        }
    //if no condition succeeds above then move to next node and continue until last node
    pCurrent=pNext;
    }
}
// delete a node is a bit different from inserting a node
void Node::deleteNode(int age)
{     
//deleting a node is much like adding one the differecne is a bit trickier
Node* pTmp=itsPtrHead;
Node* pCurrent=itsPtrHead;
Node* pNext=NULL;

//1 checking for wether age (node contains age) to be deleted does exist
for(;pTmp;pTmp=pTmp->itsPtrHead)
    {
    if(pTmp->getPdata()->getAge() == age)
        break;
    }
//if age to be deleted doesn't exist pop up a message and return
if(!pTmp)
    {
    cout<<age<<": Can't be found!\n";
    return;
    }

int NextAge=0;

for(;;)
    {
    //if age to be deleted is on the head node
    if(itsPtrHead->getPdata()->getAge() == age)
        {
        //store the next to head node
        pTmp=itsPtrHead->itsPtrHead;
        //delete head node
        delete itsPtrHead;
        //assign the head new node (node after the original head)
        itsPtrHead=pTmp;
        //decrement the count of nodes
        itsCount--;
        //exiting gracefully
        return;
        }
    //moving to next node
    pNext=pCurrent->itsPtrHead;
    NextAge=pNext->getPdata()->getAge();
    //checking next node age with age to be deleted. if they
    //match delete the next node
    if(NextAge == age)
        {
        //next node holds the target age so we want its NEXT node
        //and store it in pTmp;
        pTmp=pNext->itsPtrHead;//Next node of the target node
        //pCurrent doesn't yet hold the target age but it is the
        //previous node to target node
        //change the next node of pCurrent so that it doesn't
        //point to the target node but instead to the node right
        //after it
        pCurrent->itsPtrHead=pTmp;
        //delete the target node (holds the target age) 
        delete pNext;
        //decrement number of nodes
        itsCount--;
        //exit
        return;
        }
    //if pNext doesn't point to target node move to the next node
    //by making pCurrent points to the next node in the list
    pCurrent=pNext;
    }

}

void Node::print()const
{
Node* pTmp=itsPtrHead;
while(pTmp)
    {
    cout<<"age: "<<pTmp->getPdata()->getAge()<<endl;
    pTmp=pTmp->itsPtrHead;
    }
}
int main()
{
//note this is not a real proram just we show how things works

Data* pData=new Data(6);
Node theNode(pData);
theNode.print();
pData=new Data(19);
theNode.insertNode(pData);
pData=new Data(20);
theNode.insertNode(pData);
pData=new Data(23);
theNode.insertNode(pData);
pData=new Data(25);
theNode.insertNode(pData);
pData=new Data(30);
theNode.insertNode(pData);
pData=new Data(27);
theNode.insertNode(pData);
pData=new Data(33);
theNode.insertNode(pData);
pData=new Data(18);
theNode.insertNode(pData);

theNode.print();
cout<<endl<<endl;

//int age;
//int index;
//cout<<"Age to finde: ";
//cin>>age;
//cout<<endl;
//theNode.Find(index,age);

//cout<<age<<" : Found on index: "<<index<<endl; 

//theNode.modify(age);

//theNode.print();
int age;
cout<<"age to delete: ";
cin>>age;
cout<<endl;

theNode.deleteNode(age);
theNode.print();

cout<<endl<<endl<<endl;
return 0;
}
//modify and other member functions are not the purpose of this program
于 2015-01-03T21:26:02.960 回答
0 
void Delete()
{
    int num;
    cout<<"enter node to delete"<<endl;
    cin>>num;
    node *nodeptr=head;
    node *prev;

    if(head==0)
    {   
        cout<<"list is empty"<<endl;
    }
    else if(head->data==num)
        {
          node *t=head;
          head=head->next;
          delete t; 
        }   
        else
        {
        nodeptr=head;
        prev=head;
        while(nodeptr!=NULL)
        {
            if(nodeptr->data==num)
            {
                prev->next=nodeptr->next;
                node *tem=nodeptr->next;
                delete tem;
            }
            prev=nodeptr;
            nodeptr=nodeptr->next;
        }
        }
}
于 2020-11-09T12:51:48.210 回答