Question

I have question quite much related to this one I asked a while ago

place a value in the sorted position immediately

I wonder if you can use the same approach in that you step backward in a linked list to find the position it should be inserted into.

If it is possible how do you loop a linked list backward? I can’t figure it out because it seems not possible since it should be a double linked listed then if I’m not wrong? Anyway I’m working with singly linked list.

EDIT

I think I’m going for the look forward approach, this is what I made so far. I’m stuck at that point in how I should save the previous (key, value). Here’s the code so far of what’s done. The for-loop is used to look for the position I want to insert to. And I have peek forward, which will break in case it reach the end.

OK so far, now I want to insert the value to the correct position. It is here I’m stuck. How should it be done? Now when I insert keys: 2, 1, 0, 3, it will only print out 1, 3

struct my_list
{
  /* a pointer to the first element of the list */
  struct list_link* first;
};

struct list_link
{
   int key;                // identifies the data
   double value;           // the data stored
   struct list_link* next; // a pointer to the next data
};

struct list_link* create(int key, double value, struct list_link* next)
{
   // creates the node;
   struct list_link * new_link;
   new_link = new struct list_link;

   // add values to the node;
   new_link->key = key;
   new_link->value = value;
   new_link->next = next;

   return new_link; // Replace this, it is just to be able to compile this file
}

void list_insert(struct my_list* my_this, int key, double value)
{
   if(my_this->first == NULL)   // add if list empty
      my_this->first = create(key, value, my_this->first);   
   else
   {      
      struct my_list* curr;
      struct my_list* prev;         
      struct my_list start;

      start.first = my_this->first;
      curr = my_this;

      cout << "Too be appended: ";
      cout << key << " " << value << endl;
      for(curr->first = my_this->first; 
          key > curr->first->key; 
          curr->first = curr->first->next)
      {
         if(curr->first->next == NULL) //peek at front if empty
            break;
      }
      cout << "append here " << key << " > " << 
          curr->first->key << endl << endl;
      //perform some surgery
      if(curr->first->next == NULL)
      {     
         curr->first->next = create(key, value, my_this->first->next);
      }
      else
      {       
         curr->first = start.first; //move back to start of list
         my_this->first = create(key, value, my_this->first);
      }      
   }  
}

Answer 1

You can’t traverse a singly-linked list backward, but you can keep a pointer to the last two elements you have seen instead of just one.

So, traverse the list from the front, and keep two pointers: current, and previous. If the element you are inserting is less than current, then update previous to point to it.