c++ - 删除链表的所有其他节点

Question

我搜索了以前提出的问题，但找不到我要找的确切内容。我很好奇是否有人知道如何删除链表的每个其他节点。我有一个名为 duplicate 的函数，它将 1 2 3 转换为 1 1 2 2 3 3。删除所有其他节点就可以了，不需要比较它们或任何东西。如果有人有任何见解。请不要只发布源代码。

这是我尝试做但没有奏效的。

Node *current;
Node *undo;
for (current = front, undo = current->next->next; 
undo != NULL; current = current->next, undo = current->next->next){
    current->next = undo;
}

这将输出 1 1 2 2 3 3 3

感谢您的帮助。我可以稍后回复以澄清任何误解。

Answer 1

原代码:

for (current = front, undo = current->next->next; 
   undo != NULL; 
   current = current->next, // this moves current on before you use it's next pointer
   undo = current->next->next){
     current->next = undo;
}

修复，不释放不需要的节点(假设节点数量为偶数):

for (current = front;  current != NULL; current = current->next){
      current->next = current->next->next
}

处理奇数长度列表，并删除已删除节点的内存:

for (current = front;  current && current->next ; current = current->next){
      undo = current->next;
      current->next = current->next->next
      delete undo;
}