Enter the head node of a linked list and print the value of each node in turn from the end of the tail--5

Generally such a problem, the list is certainly not a doubly linked list is also a loop or something, that is, to give a single-linked list of the head node, and then output from the end of each node value;

If you look for the last node from the back, you will find the output and then there is no way to go back to the head, because it is just a single linked list, so you can think of it with recursion:

#include  <iostream>using namespace std;template <class T>            //defines the node of a linked list as a template class, which implements the reusability of the Code struct listnode{     T _data;    ListNode<T>* _next;}; Template <class t>listnode<t>* buy_node (T data)          //Creating a node {    listnode<t>* tmp = new listnode< t>;    tmp->_data = data;    tmp->_next =  null;    return tmp;} Template <class t>void init_list (Listnode<t>** node, t data)   Initialization of the  //list {    *node = buy_node (data);} Template <class t>void push_node (Listnode<t>*& head, t data)       //to ChainTable Insert Node {    if (head == null)     {            init_list (&head, data);         return;    }       ListNode<T>*  Tmp = head;    while (tmp->_next != null)     {            tmp = tmp->_next;     }    tmp->_next = buy_node (data);} Template <class t>void destroy_list (Listnode<t>*& head)    / /Destroy linked list {    if (head != null)     {         ListNode<T>* cur = head;         listnode<t>* tmp = head;&nBsp;       while (cur != null)          {            tmp = cur;             cur = cur->_next;             delete tmp;         }        head = NULL;     }}template <class t>void print_list (Listnode<t>* head)     //the data of the positive sequence print List {    while (head != null)     {         cout<
The above pest only to complete the problem request and did not implement the list of other operations, such as pop data and find delete insert functions, run the program can have the following results:
650) this.width=650; "src=" Http://s4.51cto.com/wyfs02/M01/7F/7A/wKioL1cgUc_zP30sAAAVdBxIpak003.png "title=" List1.png "alt=" Wkiol1cguc_zp30saaavdbxipak003.png "/>

From the above program can be known, the reverse of the list output is actually after the insertion of the node output, the first place to put the last node output, so that is the principle of LIFO, you can use the stack to achieve, from the beginning to traverse the linked list, the node one by one push_back into the stack, and then from the top of the stack to remove data, The last node of the linked list is taken, and then the pop data is then taken to the top of the stack;
The above program is a non-class variable, you can define a list class to implement, and when the program is finished running, do not manually call the destructor to free space:
#include  <iostream> #include  <vector>using namespace std;template < class t>struct listnode            // Linked list node structure body {    t _data;    listnode<t>* _next;     listnode (T data)         :_data (data)          ,_next (NULL)     {}  };template  <class T>class List                //implements a list class {public:    list ()                  //Default Constructors          : _head (NULL)     {}      list (t data)               //constructor         :_head with parameters (New ListNode <T> (data))     {}      ~list ()                //destructor, releasing the Link table node space     {            if (_head != null)          {             listnode<t>* tmp = _head;             ListNode<T>* cur = _head;             while (cur != null)              {                 tmp = cur;                cur = cur->_next;                 delete  tmp;            }             _head = NULL;         }    }    void _push (T data)                          //push data     {        if at the tail of the linked list (_head ==  null)         {             _head = new ListNode<T> (data);             return;        }         listnode<t>* tmp = _head;        while (tmp- >_next != null)             tmp =  tmp->_next;        tmp->_next = new  listnode<t> (data);     }    void print_list ()                 //positive sequence output chain list      {        listnode<t>* tmp = _head;         while (tmp != null)          {            cout<<tmp->_data< < ";   "         tmp = tmp->_next;         }        cout<< "NULL" <<endl;     }    void reverseprintlist ()                //Reverse output Chain list     {         vector<T> list;        ListNode< T>* tmp = _head;        while (Tmp != NULL)              //link table nodes into the stack in turn          {             List.push_back (tmp->_data);             tmp  = tmp->_next;        }        cout<< " Reverse print list: "<<endl;        while (!list.empty () &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;//continues to take the top element of the stack, releasing the top element of the stack          {             Cout<<list.back () <<;             list.pop_back ();        }         cout<< "NULL" <<endl;    }private:    listnode <T>* _head;}; Int main () {    list<int> list (1);     list._push (2);     list._push (3);     list._push (4);     list._push (5); &nbsP;   list._push (6);     list._push (7);     list._push (8);     list._push (9);     list.print_list ();     list. Reverseprintlist ();     return 0;}

Running the program can have the following results:
650) this.width=650; "src=" Http://s1.51cto.com/wyfs02/M01/7F/7E/wKiom1cgd8ryTugJAAAIV2OqMMY154.png "title=" List2.png "alt=" Wkiom1cgd8rytugjaaaiv2oqmmy154.png "/>


Finish
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Enter the head node of a linked list and print the value of each node in turn from the end of the tail--5