Question

代码在执行时挂起。请帮我理解这里有什么问题它在第一次遍历后停滞，甚至没有进入反向功能。无法理解这个问题：帮助我。

# include<iostream>
# include<stdlib.h>

using namespace std;

struct sLL
{
   int data;
   sLL * next;
};

void createList(sLL ** head, int n)
{
   if(n == 0)
   {
      return;
   }
   sLL * temp = (sLL *) malloc(sizeof(sLL));
   if(temp != NULL)
   {
      temp->data = rand() % 100;
      temp->next = NULL;
      *head = temp;
      //head = &(temp->next);
      createList(&(temp->next), n - 1);
   }
}

void traverse(sLL * head)
{
   cout<<"In traverse"<<endl;
   while(head) {
      cout<<head->data<<"->";
      head=head->next;
   }
   cout<<"NULL"<<endl;
   cout.flush();
   return;
} 

 sLL* reverse(sLL * head)
 {
    sLL * temp = NULL, * newNode = NULL;
    while(head) {   
       newNode = head->next; // to traverse forword
       head->next = temp;
       temp = head; // Current node value which we use in next itr as a previous node value.
       newNode = head; // assigning newNode vlaue back to head so that we can traverse forward.
    }
    return temp; // final node value will be in temp i.e the current node value.
}

int main()
{
    int n;
    sLL * head = NULL, *temp = NULL;
    cout<<"Enter the number of nodes :";
    cin>>n;
    createList(&head, n);
    cout<<"\nCreate Done"<<endl;
    traverse(head);
    cout.flush();
    cout<<"Reverse Start";
    temp = reverse(head);
    cout<<"reverse done";
    traverse(temp);
}

Answer 1

您根本不会重新分配头部值

while(head)
    //loop

这里，头指针永远不会改变。所以循环是无限的。

Answer 2

您的评论与此处的代码相矛盾：

newNode = head; // assigning newNode vlaue back to head so that we can traverse forward.

如果您执行评论所说的内容应该会更好：

head = newNode; // assigning newNode value back to head so that we can traverse forward.

代码在运行时挂起

2 个答案: