原地逆置链表,不带返回的递归方法,能实现吗?

2019-09-03 11:38:22

#include <iostream>

using namespace std;

struct ListNode
{
    int m_nValue;
    ListNode* m_pNext;
};

void add_to_linklist(ListNode** head, int e);
void show_link_list(const ListNode* head);
void recurse_reverse_linklist(ListNode** head);
ListNode* recurse_reverse_linklist(ListNode* head);

int main(){

    try
    {
        ListNode* head = NULL;
        add_to_linklist(&head, 1);
        add_to_linklist(&head, 2);
        add_to_linklist(&head, 3);

        cout << "原始链表" << endl;
        show_link_list(head);

        cout << endl << "有返回逆置,递归" << endl;
        head = recurse_reverse_linklist(head);
        show_link_list(head);
        head = recurse_reverse_linklist(head);
        show_link_list(head);
        
        cout << endl << "原地逆置,递归" << endl;
        recurse_reverse_linklist(&head);
        show_link_list(head);

        recurse_reverse_linklist(&head);
        show_link_list(head);
        
    }
    catch(const std::exception &e)
    {
        std::cout << e.what() << std::endl;
    }

    system("pause");
    return 0;
}

void recurse_reverse_linklist(ListNode** head)
{ // 无返回,原地逆置,递归
    if(*head == NULL || (*head)->m_pNext == NULL)
        return ;

    // ?????????????????????????????????
    // 此处怎样才能实现就地逆置
    // 代码不能正确逆置

    ListNode *p = (*head)->m_pNext;
    recurse_reverse_linklist(&p);
    p->m_pNext = *head;
    (*head)->m_pNext = NULL;
}

ListNode* recurse_reverse_linklist(ListNode* head)
{ // 有返回,原地逆置,递归
    if(head == NULL || head->m_pNext == NULL)
        return head;

    ListNode* newHead = recurse_reverse_linklist(head->m_pNext);

    head->m_pNext->m_pNext = head;
    head->m_pNext = NULL;

    return newHead;
}

void add_to_linklist(ListNode** head, int e)
{
    if(*head == NULL)
    {
        *head = new ListNode();
        (*head)->m_nValue = e;
        (*head)->m_pNext = NULL;
    }
    else
    {
        ListNode* p = *head;
        while(p->m_pNext)
            p = p->m_pNext;

        ListNode* pEle = new ListNode();
        pEle->m_nValue = e;
        pEle->m_pNext = p->m_pNext;
        p->m_pNext = pEle;
    }
}

void show_link_list(const ListNode* head)
{
    const ListNode* p = head;
    std::cout << "-------------------------------------------" << std::endl;

    while(p != NULL) {
        std::cout << p->m_nValue << ",";
        p = p->m_pNext;
    }
    std::cout << std::endl;
}

带返回的成功实现

不返回的逆置失败.........................链表被截断了,

输出:

原始链表

-------------------------------------------

1,2,3,

有返回逆置,递归

-------------------------------------------

3,2,1,

-------------------------------------------

1,2,3,

原地逆置,递归