如何删除链接列表中的节点？

Z时代
2024-01-10
分类：问答

这是我迄今为止，但它不是在测试文件上工作。基本上跳到别的if(cnode == preposition)如何删除链接列表中的节点？

void LinkedList::Delete(Node *PrePosition) { 
    Node *cnode = head; 
    Node *pnode = NULL; 
    while (cnode != NULL) { 
     if (cnode->value != NULL) { 
      if (pnode == NULL) { 
       //if there is not previous node 
       head = cnode->next; 
      } 
      else if(cnode == PrePosition){ 
       //if there is previous node 
       cout << endl << "Deleting: " << cnode << endl; 
       pnode->next = cnode->next; 
      } 
     } 
     else { 
      //dont delete 
      pnode = cnode; 
     } 
     cnode = cnode->next; 
    } 
}

回答：

1：从以前的节点带指针，它指向要删除

2一前一后的下一个：从以前的节点删除指针当前节点

3：删除从下一个节点的指针到当前节点（如果它是一个双向链表）

回答：

三起案件在单链表中删除的：

从codementor删除第一个节点

void delete_first() 
{ 
    node *temp=new node; 
    temp=head; 
    head=head->next; 
    delete temp; 
}

删除最后一个节点

void delete_last() 
{ 
    node *current = new node; 
    node *previous = new node; 
    current=head; 
    while(current->next != NULL) 
    { 
     previous = current; 
     current = current->next; 
    } 
    tail = previous; // if you have a Node* tail member in your LinkedList 
    previous->next = NULL; 
    delete current; 
}

删除在特定的位置（你的情况）

void LinkedList::delete_position(int pos) 
{ 
    node *current=new node; 
    node *previous=new node; 
    current=head; 
    for(int i=1; i < pos; i++) //or i = 0; i < pos-1 
    { 
     previous=current; 
     current=current->next; 
    } 
    previous->next=current->next; 
}

^^ ^^

但是，如果你的函数签名打算delete_node(Node* nodeToDelete) [介词是不是一个好名字在这种情况下]你想删除传递给函数的节点在不知道我们可以修改delete_position（在列表中的位置），像这样：

void LinkedList::delete_node(Node* nodeToDelete) 
{ 
    if (head == nodeToDelete){ 
     head = nodeToDelete->next; 
     return 
    }//else: 
    node *current=new node; 
    node *previous=new node; 
    current=head; 
    while(current != nodeToDelete) 
    { 
     previous = current; 
     current = current->next 
    } 
    previous->next = current->next; 
}