Question

我正在尝试删除双向链表中的节点，但问题是当我有列表时：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

要删除您迭代的节点，直到到达删除del节点，然后设置：伪代码：

del.prev.next = del.next

和

del.next.prev = del.prev

Answer 2

这没有任何意义：

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

因为它与

相同

del->Next = del->Next;

尝试这样的事情

// 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;
 }

注意：我修复了很多错误并尝试将其保留在您的逻辑中。我会写不同的。

双链表：删除一个节点

2 个答案: