双链表：删除节点

Question

我试图删除双向链表中的一个节点，但这里的问题是，当我有清单：7-6-5-4-3-2-1，并尝试删除3，我得到7-6-5-4-3，我无法弄清楚错误。请帮忙！

void RemoveItem(int itm) 
{   
    if (Head->item == itm) { 
     Head = Head -> Next;  
     cout<<"\nItem is at 1st node & removed\n";flag=1; 
    } 
    else if (Head->Next == NULL) {   
     cout<<" \n This is 1-node list & item not in it\n"; 
    } 
    else {     
     node *current, *del = Head; 
     while(del->Next != NULL && del->item != itm) { 
      del = del->Next; 
     } 
     current = del; 
     current->Next = del->Next; 
     del->Next->previous = current; 
     delete(del); 
     del->Next = NULL; 
     flag = 1; 
    } 
} 

Answer 1

要删除重复的节点，直到到达删除的节点德尔然后设置： 伪代码：

del.prev.next = del.next 

和

del.next.prev = del.prev 

Answer 2

这是没有意义的：

current = del; 
current->Next = del->Next; 

，因为它是一样的

del->Next = del->Next; 

尝试是这样的

// some global int flag = 0 
// some global node* Head pointing to head of list 

void RemoveItem(int itm) {   
    if (Head->item == itm) { 
    node* del = Head; 
    Head = Head -> Next; 
    delete(del);  
    cout<<"\nItem is at 1st node & removed\n"; 
    flag=1; 
    return; 
    } 
    if (Head->Next == NULL) {    
    cout<<" \n This is 1-node list & item not in it\n"; 
    return; 
    }     
    node* current = Head; 
    node* del = Head; 

    while(del->Next != NULL && del->item != itm) { 
    del = del->Next; 
    } 

    // reached with del->Next == NULL || del->item == itm 
    current = del->previous; 
    if (del->Next == NULL) { 
    // at end of list 
    if (del->item != item) { 
     // item not found 
     // todo: signal error 
     return; 
    } 
    current->Next = NULL; 
    } else { 
    del->Next->previous = current; 
    current->Next = del->Next; 
    } 
    delete(del); 
    flag = 1; 
} 

注：我定了许多错误，并试图保持它的逻辑中。我会写它与众不同。