我有一个编写反转函数的赋值,用于反转最多n个块的链接双向链表。我首先从forloop中获取大小的端点,然后我将tartpoint和端点发送到外部函数以反转它们。外部功能成功地尊重了头部和尾部,但是我正在进行分裂,同时扭转一个给定的大小。我需要帮助解决出错的问题?
反向功能;
/**
* Helper function to reverse a sequence of linked memory inside a List,
* starting at startPoint and ending at endPoint. You are responsible for
* updating startPoint and endPoint to point to the new starting and ending
* points of the rearranged sequence of linked memory in question.
*
* @param startPoint A pointer reference to the first node in the sequence
* to be reversed.
* @param endPoint A pointer reference to the last node in the sequence to
* be reversed.
*/
template <class T>
void List<T>::reverse( ListNode * & startPoint, ListNode * & endPoint )
{
if( (startPoint == NULL) || (endPoint == NULL) || (startPoint == endPoint))
{ return; }
ListNode * curr = startPoint;
ListNode * nexter = NULL;
ListNode * prever = endPoint->next;
while( curr != NULL)
{
nexter = curr->next;
curr->next = prever;
prever = curr;
curr = nexter;
prever->prev = curr;
}
// now swap start and end pts
nexter = startPoint;
startPoint = endPoint;
endPoint = nexter;
}
现在给出sze的反向函数,它应该使用上面的函数;
/**
* Reverses blocks of size n in the current List. You should use your
* reverse( ListNode * &, ListNode * & ) helper function in this method!
*
* @param n The size of the blocks in the List to be reversed.
*/
template <class T>
void List<T>::reverseNth( int n )
{
if(n == 0)
return;
ListNode * startPoint = head;
ListNode * endPoint = head;
ListNode * save = NULL;
for(int i = 0; i< n; i++) // need to get endpoint at n
{
endPoint = endPoint->next;
}
reverse(startPoint, endPoint);
}
gdb会输出一些奇怪的东西,可能是因为在图像上工作的函数失败了;
Program received signal SIGINT, Interrupt.
0x000000000040dcab in __distance<List<RGBAPixel>::ListIterator> (__first=..., __last=...) at /class/cs225/llvm/include/c++/v1/iterator:488
488 for (; __first != __last; ++__first)
(gdb) q
A debugging session is active.
Inferior 1 [process 31022] will be killed.
答案 0 :(得分:1)
我需要回复此问题,因为我不知道是否可以在评论中执行格式化代码。
那就是说,你在这方面的工作太过努力了:
ListNode * curr = startPoint;
while( curr != NULL)
{
ListNode * temp = curr->next;
curr->next = curr->prev;
curr->prev = temp;
curr = temp;
}
应该让你到达你想去的地方。