如何刪除鏈接列表中的節點？

Question

這是我迄今爲止，但它不是在測試文件上工作。基本上跳到別的if(cnode == preposition)

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; 

    } 

} 

Answer 1

1：從以前的節點帶指針，它指向要刪除

2一前一後的下一個：從以前的節點刪除指針當前節點

3：刪除從下一個節點的指針到當前節點（如果它是一個雙向鏈表）

Answer 2

三起案件在單鏈表中刪除的：

1. 從codementor刪除第一個節點

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

2. 刪除最後一個節點

   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; 
   } 
   

3. 刪除在特定的位置（你的情況）

   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，如果你的函數簽名打算delete_node(Node* nodeToDelete) [介詞是不是一個好名字在這種情況下]你想刪除傳遞給函數的節點在不知道我們可以修改delete_position（在列表中的位置），像這樣：

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; 
} 

另外在你的原代碼，如果它跳過你所提到的線，pnode總是空當cnode中有一個非空值。