反向打印链表

Question

my English is not very well, but hope u guys can understand me what i'm trying to say. So this is the code for linked list, and after run the program and add the information, it can be print at printListStart().Now i having trouble for writing a code at printListEnd(),and i want to show the code from end(reverse).

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <conio.h>


//function prototype
void addToStart();      //add node to beginning of linked list
void addToEnd();        //add node to end of linked list
void removeNodeAt();    //remove node that matches element entered
void printListStart();      //print nodes from start
void printListEnd();        //print node from end

void startlist();   //create NULL list
void menu();        //selection

//global variables
int option, number;
char name[20], gender[10],address[50],description[50];


//declare structure for node
struct node
{
    char customer_name[20];
    int customer_number;
    char gender_[10];
    char customer_address[50];
    char order_description[50];
    struct node *next;

}*newnode, *list, *prev, *temp, *tmpdisplay;

void main()
{

    startlist();    //function call to create empty list

    do
    {

        menu();     //function call to show menu

        switch (option)
        {
        case 1: system("cls"); addToStart(); break;
        case 2: system("cls");  addToEnd(); break;
        case 3: system("cls");  removeNodeAt(); break;
        case 4: system("cls");  printListStart(); break;
        case 5: system("cls"); printListEnd(); break;
        case 6: exit(0);
        default:
            printf("Invalid Option");
            getch();
        }

    } while (option != 6);


}//end main


void startlist()
{
    list = NULL;        //create empty list 
}

void menu()
{

    printf("***LINKED LIST***\n\n");
    printf("   1. Add New Node At Start\n");
    printf("   2. Add New Node At End\n");
    printf("   3. Remove Node\n");
    printf("   4. Print Linked List From Start\n");
    printf("   5. Print Linked List From End\n");
    printf("   6. Quit\n");

    printf("\nSelect a task: ");        //allow user to select choice
    scanf("%d", &option);
}

void addToStart()
{
    newnode = (struct node*) malloc(sizeof(struct node));   //allocates memory space for new node

    printf("Enter the customer name:\n");
    scanf("%s", &name);
    printf("Enter then customer number:\n");
    scanf("%d", &number);
    printf("Enter the Oder Description:\n");
    scanf("%s", &description);
    printf("Enter the Gender:\n");
    scanf("%s", &gender);
    printf("Enter the Customer Address:\n");
    scanf("%s", &address);

    newnode->customer_number = number;
    strcpy(newnode->customer_name, name);       //using stringcopy to copy name to customer_name in node
    strcpy(newnode->order_description, description);     //using stringcopy to copy transdes to transaction_description in node
    strcpy(newnode->gender_, gender);
    strcpy(newnode->customer_address, address);
    newnode->next = NULL;   //set node pointer to NULL



    if (list == NULL)
        list = newnode; //if list is empty, node is assigned to list
    else
    {
        newnode->next = list;   //if list not empty, newnode pointer equals to list first node
        list = newnode;     //assign newnode to list, newnode is at the start of the list
    }
}


void addToEnd()
{
    newnode = (struct node*) malloc(sizeof(struct node));   //allocate new memory space for new node

    printf("Enter the customer name:\n");
    scanf("%s", &name);
    printf("Enter then customer number:\n");
    scanf("%d", &number);
    printf("Enter the Oder Description:\n");
    scanf("%s", &description);
    printf("Enter the Gender:\n");
    scanf("%s", &gender);
    printf("Enter the Customer Address:\n");
    scanf("%s", &address);

    newnode->customer_number = number;
    strcpy(newnode->customer_name, name);
    strcpy(newnode->order_description, description);
    strcpy(newnode->gender_, gender); 
    strcpy(newnode->customer_address, address);
    newnode->next = NULL;

    if (list == NULL)
        list = newnode; //if list is empty, assign newnode to list as first node
    else
    {
        temp = list;        //list not empty, assign temp as list
        while (temp->next != NULL) //while pointer does not point to NULL/empty
        {
            temp = temp->next;      //move to subsequent node
        }
        temp->next = newnode;       //loop exits when last node is reached, last node's pointer points to newnode 
    }
}

void removeNodeAt()
{
    printf("Enter customer number to delete: \n");
    scanf("%d", &number);

    if (list == NULL)       //check if list is empty
        printf("\n\nLIST IS EMPTY\n\n");

    //if list not empty, match number to cust_no in first node
    else if (number == list->customer_number)
    {

        list = list->next;  //match found, first node is skipped (deleted)
    }
    else //match not found in first node, move to subsequent nodes  
    {
        temp = list;    //assign temp as list
        while (temp->customer_number != number)
        {
            //if match not found
            prev = temp;        //prev is pointing to linked list
            temp = temp->next;//temp is pointing to next node
        }

        printf("Node deleted:");
        printf("\n%s\n", prev->customer_name);
        printf("%d\n", prev->customer_number);
        printf("%s\n\n", prev->gender_);
        prev->next = prev->next->next;  //match found, skip/jump the node (delete)
    }
}

void printListStart()
{
    if (list == NULL)
        printf("\n\nLIST IS EMPTY\n\n");
    else
    {
        tmpdisplay = list;
        while (tmpdisplay != NULL)
        {
            printf("\n%s\n", tmpdisplay->customer_name);
            printf("%d\n", tmpdisplay->customer_number);
            printf("%s\n", tmpdisplay->gender_);
            printf("%s\n", tmpdisplay->order_description);
            printf("%s\n", tmpdisplay->customer_address);
            tmpdisplay = tmpdisplay->next;
        }
    }
}


void printListEnd()
{

}

Answer 1

Simplest way is to create a doubly-linked list, where each node has a next and previous pointer, where next points to next node (like you are doing now) and previous points to the previous node. You will also need pointers to the front and end of the list. To print it in reverse, start with pointer to last node and follow the previous pointers instead of next pointers.

struct node
{
    char customer_name[20];
    int customer_number;
    char gender_[10];
    char customer_address[50];
    char order_description[50];
    struct node *next;
    struct node *prev;
}*newnode, *list, *prev, *temp, *tmpdisplay, *listend;

When adding a node to the end, prev for the new node to listend, and set listend to the new node.

Answer 2

If you cannot use a doubly-linked list for some reason, your choices are to use a nested loop, or to use recursion. Both of these solutions are horrible. Here's the nested loop solution, which is horrible because it runs in O(N^2) time:

void printListEnd()
{
    struct node *end;

    if (list == NULL)
        printf("\n\nLIST IS EMPTY\n\n");
    else
    {
        end = NULL;
        while (end != list)
        {
            tmpdisplay = list;
            while (tmpdisplay->next != end)
                tmpdisplay = tmpdisplay->next;
            printf("\n%s\n", tmpdisplay->customer_name);
            printf("%d\n", tmpdisplay->customer_number);
            printf("%s\n", tmpdisplay->gender_);
            printf("%s\n", tmpdisplay->order_description);
            printf("%s\n", tmpdisplay->customer_address);
            end = tmpdisplay;
        }
    }
}

Here's the recursive solution, which is horrible because it wastes stack space in proportion to the length of the list:

static void printListEndRecurse(struct node *list)
{
    if (list->next)
        printListEndRecurse(list->next);
    tmpdisplay = list;
    printf("\n%s\n", tmpdisplay->customer_name);
    printf("%d\n", tmpdisplay->customer_number);
    printf("%s\n", tmpdisplay->gender_);
    printf("%s\n", tmpdisplay->order_description);
    printf("%s\n", tmpdisplay->customer_address);
}

void printListEnd()
{
    if (list == NULL)
        printf("\n\nLIST IS EMPTY\n\n");
    else
        printListEndRecurse(list);
}

Answer 3

Sorry but I immediately thought to the recursion. Why don't give a chance to this fantastic tecnique? Let me try:

    void printListEnd(struct node *myList)
    {
        if (myList != NULL) {
            printListEnd(myList->next);
            printf("\n%s\n", myList->customer_name);
            printf("%d\n", myList->customer_number);
            printf("%s\n", myList->gender_);
            printf("%s\n", myList->order_description);
            printf("%s\n", myList->customer_address);
        }
    }

To call the function you need to pass your variable list (the root node of your list).

Answer 4

After accept answer

Allocate an array to store the link addresses. Then print the array elements in reverse order.
O(n) time.

  // Find size
  size_t n = 0;
  struct node *p = myList;
  while (p) {
    p = p->next;
    n++;
  }

  // Build array
  struct node **list = malloc(sizeof *list * n);
  if (list) {
    struct node **walker = list;
    p = myList;
    while (p) {
      *walker = p; 
      walker++;
      p = p->next;
    }

    // Print list
    while (walker != list) {
      walker--;
      print_list(*walker);
    }

    free(list);
    list = NULL;
  }

This approach does not make use of functional recursion.