问题描述
我正在尝试使用递归来反向链接列表。我以前已经编写了代码来使用一个简单的while循环来做到这一点，并且效果很好。我不知道为什么我的递归代码不起作用。
#include<iostream>
using std::cout;
using std::endl;
struct node
{
int data;
node* next;
};
class LinkedLists
{
private:
node* head;
node* temp;
public:
LinkedLists() // default constructor
{
head = NULL;
temp = NULL;
}
void add_data(int d)
{
node* new_node = new node; // create a pointer to the new node
new_node->next = NULL;
new_node->data = d;
if (head == NULL)
{ head = new_node;
temp = head;
}
else
{
temp = head;
while(temp->next != NULL)
{
temp = temp->next;
}
temp->next = new_node; // final node Now points to the new_node
}
}
void print_list()
{
temp = head;
while(temp!=NULL)
{
std::cout<<temp->data<<" ";
temp = temp->next;
}
}
void reverse()
{
// reverse a linked list
node* prev_node;
node* next_node;
node* temp_ptr;
prev_node = NULL;
temp_ptr = head;
next_node = temp_ptr->next;
while(next_node != NULL)
{
temp_ptr->next = prev_node;
prev_node = temp_ptr;
temp_ptr = next_node;
next_node = temp_ptr->next;
}
temp_ptr->next = prev_node;
head = temp_ptr;
}
void repeat(node* prev_node,node* temp_ptr,node* next_node)
{
temp_ptr->next = prev_node;
prev_node = temp_ptr;
temp_ptr = next_node;
if (next_node != NULL)
{
next_node = temp_ptr->next;
repeat(prev_node,temp_ptr,next_node);
}
head = temp_ptr;
}
void recursive_reverse()
{
node* prev_node;
node* next_node;
node* temp_ptr;
prev_node = NULL;
temp_ptr = head;
next_node = temp_ptr->next;
repeat(prev_node,next_node);
}
};
int main()
{
LinkedLists l; // create a linked list object
l.add_data(110);
l.add_data(140);
l.add_data(101);
l.add_data(140);
l.add_data(101);
l.add_data(140);
l.add_data(101);
l.add_data(120);
cout<<endl;
l.print_list();
l.reverse();
cout<<endl;
l.print_list();
l.recursive_reverse();
cout<<endl;
l.print_list();
}
输出：
110 140 101 140 101 140 101 120
120 101 140 101 140 101 140 110
101 120
预期输出：
110 140 101 140 101 140 101 120
120 101 140 101 140 101 140 110
110 140 101 140 101 140 101 120
解决方法
您的代码是多余的，您不需要那么多局部变量和函数参数即可实现递归。也在这部分中：
if (next_node != NULL)
{
next_node = temp_ptr->next;
repeat(prev_node,temp_ptr,next_node);
}
head = temp_ptr;
您似乎在nullptr时将head分配给了next_node == nullptr。
更正的版本：
void repeat(node* previous,node* current)
{
node* next_node = current->next;
current->next = previous;
if (next_node == nullptr)
{
head = current;
return;
}
repeat(current,next_node);
}
void recursive_reverse()
{
repeat(nullptr,head);
}
除此之外，您还会发生内存泄漏，因为您从未delete分配了内存。实现析构函数并将您的副本构造函数/赋值明确标记为deleted（或者，如果有时间的话，也可以实现它们）：
LinkedLists(const LinkedLists& other) = delete;
为避免进一步的错误并遵守the rule of three（或五个或零）。
,对于初学者来说，不清楚为什么链表的类具有复数名称。
class LinkedLists
^^^
更自然地将其命名为LinkedList而不用's'结尾。
数据成员temp是冗余的，应将其删除。相反，您可以在成员函数中用此数据成员替换局部变量。
结构node应该是类LinkedList的私有成员。
我没有仔细研究您所有的函数实现，但是可以按照以下演示程序中所示的以下方式定义函数recursive_reverse。
您在这里。
#include <iostream>
#include <functional>
class LinkedList
{
private:
struct node
{
int data;
node* next;
} *head = nullptr;
public:
LinkedList() = default;
// These two special member functions you can implement yourself
LinkedList( const LinkedList &) = delete;
LinkedList & operator =( const LinkedList & ) = delete;
~LinkedList()
{
clear();
}
void clear()
{
while ( head )
{
delete std::exchange( head,head->next );
}
}
void add_data( int d )
{
node **current = &head;
while ( *current ) current = &( *current )->next;
*current = new node { d,nullptr };
}
friend std::ostream & operator <<( std::ostream &os,const LinkedList &list )
{
for ( const node *current = list.head; current != nullptr; current = current->next )
{
os << current->data << " -> ";
}
return os << "null";
}
void recursive_reverse()
{
if ( head != nullptr && head->next != nullptr )
{
node *current = head;
head = head->next;
recursive_reverse();
current->next->next = current;
current->next = nullptr;
}
}
};
int main()
{
LinkedList list; // create a linked list object
const int N = 10;
for ( int i = 0; i < N; i++ )
{
list.add_data( i );
}
std::cout << list << '\n';
list.recursive_reverse();
std::cout << list << '\n';
list.recursive_reverse();
std::cout << list << '\n';
return 0;
}
程序输出为
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
9 -> 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> 0 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
如果要对成员函数recursive_reverse使用辅助函数，则该函数应为私有静态成员函数。列表的用户不得直接调用它。
这是一个演示程序。
#include <iostream>
#include <functional>
class LinkedList
{
private:
struct node
{
int data;
node* next;
} *head = nullptr;
static void repeat( node *previous,node * &head )
{
if ( head )
{
node *current = head;
if ( head->next )
{
head = head->next;
repeat( current,head );
}
current->next = previous;
}
}
public:
LinkedList() = default;
~LinkedList()
{
clear();
}
void clear()
{
while ( head )
{
delete std::exchange( head,const LinkedList &list )
{
for ( const node *current = list.head; current != nullptr; current = current->next )
{
os << current->data << " -> ";
}
return os << "null";
}
void recursive_reverse()
{
repeat( nullptr,head );
}
};
int main()
{
LinkedList list; // create a linked list object
const int N = 10;
for ( int i = 0; i < N; i++ )
{
list.add_data( i );
}
std::cout << list << '\n';
list.recursive_reverse();
std::cout << list << '\n';
list.recursive_reverse();
std::cout << list << '\n';
return 0;
}
程序输出与上图相同
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null
9 -> 8 -> 7 -> 6 -> 5 -> 4 -> 3 -> 2 -> 1 -> 0 -> null
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9 -> null