Segvert同時反轉給定大小的鏈接列表

Question

我有一個任務來寫一個反轉函數來反轉鏈接雙向鏈表，最多n個塊。我首先從forloop中獲取端點的大小，然後我將tartpoint和端點發送到外部函數以反轉它們。外部功能成功地恢復了頭部和尾部，但是我在劃轉時扭轉了給定的尺寸。我需要一些幫助解決問題？

反轉功能;

/** 
* 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; 

}

現在給定事反向功能，它應該使用上述功能;

/** 
* 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輸出一些奇怪的東西，或許給函數的圖像之後是失敗的工作;

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. 

Answer 1

我需要把這個回覆，因爲我不知道是否有可能在評論中做格式化代碼。

這就是說，你在這方面的工作方式太辛苦：要成爲

ListNode * curr = startPoint; 
while(curr != NULL) 
{ 
    ListNode * temp = curr->next; 
    curr->next = curr->prev; 
    curr->prev = temp; 
    curr = temp; 
} 

應該得到你。