【发布时间】：2021-03-02 02:27:55
【问题描述】：
现在，我正在尝试为双向链表编写反向方法。这不仅是您从头到尾反转的方法之一，而是从一个点开始并在开始后的一个点结束的列表的一部分。该方法接受称为startPoint 和endPoint 的ListNode 指针引用，它们应该指向链接内存中链的新起点和终点。序列开始之前的ListNode 的next 成员（所以startPoint->prev）应该指向新的开始，而序列结束之后的ListNode 的prev 成员（所以@987654330 @) 应该指向新的结束。
我应该指出的一个关键方面是我们可能不分配任何新的ListNodes。
在我的 List.h 文件中，我有一个私有类 ListNode。我将它与我的 List 类的其余部分一起提供：
template <class T>
class List {
private:
class ListNode {
public:
ListNode();
ListNode(const T & ndata);
ListNode* next;
ListNode* prev;
const T data;
};
public:
List();
List(const List<T>& other);
List<T>& operator=(const List<T>& rhs);
int size() const;
void print(ostream& os = std::cout) const;
bool empty() const;
~List();
void insertFront(const T& ndata);
void insertBack(const T& ndata);
void reverse();
void tripleRotate();
List<T> split(int splitPoint);
private:
ListNode* head_;
ListNode* tail_;
int length_;
void _copy(const List<T>& other);
void _destroy();
/**
* 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.
*/
void reverse(ListNode*& startPoint, ListNode*& endPoint);
};
在我的 List.hpp 文件中，我有反向方法本身。我不知道如何准确地写它。这是我到目前为止所拥有的（显然不起作用）：
template <typename T>
void List<T>::reverse(ListNode *& startPoint, ListNode *& endPoint) {
if (startPoint == endPoint) {
return;
}
//startPoint should point at the new start, endPoint should point at the new end.
ListNode* current = startPoint;
ListNode* before_start_point = startPoint->prev;
ListNode* after_end_point = endPoint->next;
while (current != after_end_point) {
ListNode* temp = current->next;
current->next = current->prev;
current->prev = temp;
if (temp == endPoint) {
endPoint = startPoint;
startPoint = current;
}
current = temp;
}
}
List.hpp 文件的其余部分：
template <class T>
List<T>::List() {
head_ = NULL;
tail_ = NULL;
length_ = 0;
}
template <typename T>
void List<T>::_destroy() {
ListNode* current = head_;
while (current != NULL) {
ListNode* temp = current->next;
delete current;
current = temp;
}
}
template <typename T>
void List<T>::insertFront(T const & ndata) {
ListNode* newNode = new ListNode(ndata);
//Case where there the list is empty
if (head_ == NULL) {
head_ = newNode;
tail_ = newNode;
newNode->next = NULL;
newNode->prev = NULL;
}
else {
newNode->next = head_;
newNode->prev = NULL;
head_->prev = newNode;
head_ = newNode;
}
length_++;
}
template <typename T>
void List<T>::insertBack(const T & ndata) {
ListNode* newNode = new ListNode(ndata);
if (tail_ == NULL) {
head_ = newNode;
tail_ = newNode;
newNode->next = NULL;
newNode->prev = NULL;
}
else {
newNode->prev = tail_;
newNode->next = NULL;
tail_->next = newNode;
tail_ = newNode;
}
length_++;
}
template <typename T>
typename List<T>::ListNode* List<T>::split(ListNode* start, int splitPoint) {
//There will be splitPoint number of nodes remaining in the current list
ListNode* curr = start;
if (splitPoint == 0) {
return curr;
}
//Takes you to the start of the new list
for (int i = 0; i < splitPoint && curr != NULL; i++) {
curr = curr->next;
}
if (curr != NULL) {
curr->prev->next = NULL;
curr->prev = NULL;
}
//Return the head of the new sublist
return curr;
}
template <typename T>
void List<T>::tripleRotate() {
if (length_ < 3) {
return;
}
else {
int third_element_counter = 1;
bool first_rotation = true;
int divisible_by_three = length_ % 3;
ListNode* current = head_;
while (current != NULL) {
if (third_element_counter != 3) {
third_element_counter++;
}
else {
ListNode* first = current->prev->prev;
ListNode* temp_first_prev = first->prev;
ListNode* second = current->prev;
ListNode* temp_current = current;
ListNode* temp_current_next = current->next;
second->prev = temp_first_prev;
if (temp_first_prev != NULL) {
temp_first_prev->next = second;
}
if (temp_current_next != NULL) {
temp_current_next->prev = first;
}
current->next = first;
first->next = temp_current_next;
first->prev = temp_current;
if (first_rotation) {
head_ = second;
first_rotation = false;
}
if (divisible_by_three == 0) {
tail_ = first;
}
current = first;
third_element_counter = 1;
}
current = current->next;
}
}
}
template <typename T>
void List<T>::reverse() {
reverse(head_, tail_);
}
感谢任何形式的帮助。
【问题讨论】：
-
您没有提供足够的代码。
List类是什么？ -
@MysteriousUser 刚刚编辑了我的帖子。您现在应该可以看到 List 类了。
-
@HughJass24 能否请您提供
List方法的实现？所以我们可以完全测试它而无需实现它们。 -
@MysteriousUser 刚刚添加了实现。我事先测试了所有其他功能，它们都可以工作。但是，反向功能不起作用。
标签： c++ linked-list doubly-linked-list