Question

我正在编写一个函数，在与head2关联的列表中发生变量位置后，输入与head1关联的链接列表。但是，我一直在获得核心转储：

void mergeLists(ListNode *head1, ListNode *head2, const int &location){
  ListNode *tail1, *tail2, *run;
  tail1=head1;
  tail2=head2;
  if(head1->pointer_Next!=nullptr){
    while(tail1->content!=location){
      tail1=tail1->pointer_Next;
    }
    if(head2->pointer_Next!=nullptr){
      while(tail2->pointer_Next!=nullptr){
        run=tail1->pointer_Next;
        tail1->pointer_Next=new ListNode;
        tail1=tail1->pointer_Next;
        tail1->content=tail2->content;
        tail1->pointer_Next=run;
        tail2=tail2->pointer_Next;
      } 
    }
  }
  delete tail1;
  delete tail2;
  delete run;
}

第12行的操作是否有违法行为？我是通过GDB运行的，我很确定这就是问题所在。我已经尝试将指针设置为nullptr旁边但它产生相同的结果。有没有人知道核心转储在哪里发生？

Answer 1

您的代码存在很多问题，即使没有调试也能看到。请发布测试用例+错误+ ListNode定义。

void mergeLists(ListNode *head1, ListNode *head2, const int &location){
  ListNode *tail1, *tail2, *run;
  tail1=head1;
  tail2=head2;
  if(head1->pointer_Next!=nullptr){ <------ What if head1 is nullptr ?
    while(tail1->content!=location){  <---- What if tail1 is nullptr ?
      tail1=tail1->pointer_Next;      <---- What if tail1->pointer_Next is nullptr ?
    }
    if(head2->pointer_Next!=nullptr){  <--- What if head2 is nullptr ?
      while(tail2->pointer_Next!=nullptr){ <--- What if tail2 is nullptr ?
        run=tail1->pointer_Next;
        tail1->pointer_Next=new ListNode;
        tail1=tail1->pointer_Next;
        tail1->content=tail2->content;
        tail1->pointer_Next=run;
        tail2=tail2->pointer_Next;
      } 
    }
  }
  delete tail1; <---- Why do you delete tail1 , which is Node in the list
  delete tail2; <---- Why do you delete tail2 , which is Node in the list
  delete run;
}

是否可以将新的ListNode分配给链接列表的现有节点？

1 个答案: