单链表前插法

其实根本没有什么前插法,这是一个坑!!!

因为获取不到前一个结点的next指针,故,实现方法为:向当前结点后面插入一个新的结点,交换两个结点的值
c++代码实现如下:

#include 
#include 
using namespace std;

template < typename T>
class Node{
    public:
        ~Node();
        Node<T>* next{nullptr};
        T data;
};

template <typename T>
Node<T>::~Node()
{
    next = nullptr;
}

template <typename T>
void insert(Node<T>* s, T data)
{
    if(s == nullptr) return;                                                                     
    Node<T>* p = new Node<T>();
    p->next = s->next;
    s->next = p;
    p->data = s->data;
    s->data = data;
}

template <typename T>
void erase(Node<T>* s)
{
    if(s == nullptr || s->next == nullptr) return;
    Node<T>* p = s->next;
    s->data = std::move(p->data);
    s->next = p->next;
    delete p;
    p = nullptr;
}

int main()
{
    auto head = new Node<int>();
    head->data = -1;
    auto t = new Node<int>();
    head->next = t;
    t->data = 4;
    insert(t, 8);
    insert(t, 9);
    insert(t, 5);
    insert(t, 1);
    Node<int>* s = head;
    while(s)
    {
        cout<< s <<":"<<s->data <<" ";
        s = s->next;
    }
    cout << endl;
    Node<int>* p = head->next;
    erase(p);
    Node<int>* x = head;
    while(x)
    {
        cout<< x <<":"<<x->data <<" ";
        x = x->next;                                                                             
    }
    cout << endl;
    return 0;
}

结果:

0xff4010:-1 0xff4030:1 0xff40b0:5 0xff4090:9 0xff4070:8 0xff4050:4 
0xff4010:-1 0xff4030:5 0xff4090:9 0xff4070:8 0xff4050:4

附上一篇打脸文章
https://blog.csdn.net/helloworld_ptt/article/details/86522516