Question

我需要做的是复制已填充这些值的链表list1：

并将它们与函数CreateMenuList（）粘贴到另一个名为list2的列表中，以便list2的每个元素都具有与特定数字匹配的list1成员，ViewAllMenu（）的输出应为：

现在我刚刚写了这个基本概念：

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

struct Order { 
    char code[8];
    int menu;
};

typedef struct NodeO {
    struct Order order;
    struct NodeO *next;
} TNode;

typedef TNode * NodeO;

struct Menu {
    int code;
    NodeO orders_list;
};

typedef struct NodeM {
    struct Menu menu;
    struct NodeM * next;
} TNodeM;

typedef TNodeM * NodeM;

//this function is for creating nodes for list1
NewOrder(struct Order p, NodeO * pp)
{
    NodeO temp;

    temp = malloc(sizeof(struct NodeO));
    temp->order = p;
    temp->next = *pp;

    *pp = temp;

}

void main()
{
    NodeO list1 = NULL;
    NodeM list2 = NULL;

    //supposing list1 is already filled
    CreateMenuList(list1,list2);
    ViewAllMenu(lista2);
}

假设我们已经使用使用stdin调用list1的函数填充NewOrder，函数CreateMenuList应该做些什么来修改list2并创建n会员存储与菜单匹配的所有订单？ ViewAllMenu并不是真正的问题，因为它应该是printf的简单while循环。

Answer 1

这些是所需的所有功能：

void Load(NodeO * p)
{
  FILE *f;
  struct Order s;
  char * buffer = malloc(sizeof(struct Order));

  if(!(f=fopen("test.txt", "r")))
   {
    perror("Errore");
    exit(EXIT_FAILURE);
   }

  while(fgets(buffer,sizeof(struct Order), f))
  {
      if(sscanf(buffer, "%s%d", s.code, &s.menu) == 2 )
      {
        NewOrder(s,p);
      }
  }
  free(buffer);
  fclose(f);
}

void InsertOrder(NodeM * pp, struct Order p)
{
  NodeM  s;

    s = *pp;

    while(s != NULL && (s->menu.code != p.menu))
    s = s->next;

    if( s == NULL)
    {
    s = (NodeM)malloc(sizeof(TNodeM));
           s->next = *pp;
               *pp = s;
      s->menu.code = p.menu;
      s->menu.list = NULL;
    }

    NewOrder(p, &s->menu.list);
}

void CreateMenuList(NodeO p, NodeM * pp)
{
   while(p != NULL)
   {
      InsertOrder(pp, p->order);
      p = p->next;
   }
}



void ViewAllMenu(NodeM p)
{
  NodeO s;
  while(p != NULL)
  {
    s = p->menu.list;
    printf("%d\n", p->menu.code);
    while(s != NULL)
    {
      printf("\t%s\n", s->order.code);
      s = s->next;
    }
   p = p->next;
  }

}

如何按顺序将元素从一个链表移动到另一个链表

1 个答案: