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reversing a linklist

Hi how do I reverse a linklist, I tried making a reverse function based on the idea that I should swap the pointers. But wasn't able to implement it properly.

I actually want to reverse the same list instead of creating a templist and then adding to it the values of the original list and then destroying the list.

Last thing I need to know is how to overload the = operator which sets one linklist equal to another

------------------------------
#include "iostream.h"

struct link
{

      int x;
      link *next;

};

class linklist
{

      link * first;
public:
      linklist()
      {
            first=NULL;
      }

      void additem(int d);
      void display();
      void reverse();
};

void linklist::additem(int d)
{
 
      link *nextlink = new link;
      nextlink->x=d;
      nextlink->next=first;
      first=nextlink;



}

void linklist::display()
{

      link *current=first;

      while (current !=NULL)
      {
      
            cout << current->x <<endl;
            current=current->next;
      }

}


void linklist::reverse()

{
      link *current=first;
      link *temp=NULL;
      link *count;

      //int temp=0;

      cout <<"first= " << first <<endl;

      while(count !=NULL)
      {
      
            temp=current;
            current=current->next;
            current->next=temp;
            count=current->next->next;
      }

}


void main()
{

      linklist li;

      li.additem(5);
      li.additem(43);
      li.additem(86);

      li.reverse();
      
      li.display();


}
0
anshuma
Asked:
anshuma
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2 Solutions
 
efnCommented:
An algorithm for reversing a linked list:

while the source list is not empty
{
  remove the first element from the input list
  add the element to the beginning of the output list
}

An algorithm for assigning to a linked list:

clear the target list
start at the beginning of the source list
while not at the end of the source list
{
  make a copy of the current source element
  add the copy to the end of the target list
  advance to the next source element
}

Or were you looking for the syntax for declaring an assignment operator?
0
 
Jaime OlivaresSoftware ArchitectCommented:
This is my first attempt (not tested):

link *invert_list(link *node, link *prev)
{
   if (node->next)
         invert(node->next, node);
   else
         node->next = prev;
   return node;
}

0
 
Jaime OlivaresSoftware ArchitectCommented:
Ok, I got it. I'll send full code. Please advice if any problem.

#include <iostream>

using namespace std;

struct link
{
     int x;
     link *next;
};

link *invert_list(link *node, link *prev)
{
    link *last;

   if (node->next)
         last = invert_list(node->next, node);
   else
         last = node;
         
   node->next = prev;
   return last;
}


class linklist
{
     link * first;
public:
     linklist()
     {
          first=NULL;
     }

     void additem(int d);
     void display();
     void reverse();
};

void linklist::additem(int d)
{
     link *nextlink = new link;
     nextlink->x=d;
     nextlink->next=first;
     first=nextlink;
}

void linklist::display()
{
     link *current=first;

     while (current !=NULL)
     {
          cout << current->x <<endl;
          current=current->next;
     }
}

void linklist::reverse()
{
    first = invert_list(first, NULL);
}

int main()
{
     linklist li;

     li.additem(5);
     li.additem(43);
     li.additem(86);

     cout << "Normal" << endl;
     li.display();
     li.reverse();
     cout << "Reversed" << endl;
     li.display();
   
     return 0;
}
0
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efnCommented:
>  Please advice if any problem.

With a sufficiently large list (11,000 was enough when I tested), the program blows up with a stack overflow.  This might be considered a problem.
0
 
Jaime OlivaresSoftware ArchitectCommented:
Memory comsumsion and performance always are compromised. Just choose the right for you.
0
 
Jaime OlivaresSoftware ArchitectCommented:
quicksort for a huge  list or array also can fail due its recursive nature. If you plan to reverse a huge linked list, then you can divide it in portions, inverse every portion and link all them again. But I would do it only if strictly necessary.
0
 
bcladdCommented:
My argument against jaime_olivares' argument is that in this case there is an equally easy to write ITERATIVE solution. Quicksort is much harder to rewrite iteratively.

Try:

void reverse()
{
  node* prev = NULL, * curr = head;
 
  while (curr != NULL) {
    node * next = curr->next;
    curr->next = prev;
    prev = curr;
    curr = next;
  }
  head = prev;
}


-bcl
0
 
efnCommented:
bcladd's solution, which is equivalent to the one I outlined above, is preferable because it requires storing 3 pointers rather than 2n pointers.  I consider this more significant than the fact that it is iterative, although that also saves memory.  From my experience trying to code this in job interviews, I think bcladd's solution, while not conceptually difficult, is actually harder to write than jaime_olivares's solution.
0
 
bcladdCommented:
No question the recursive solution is the easier one to think through. It is not hard to work it around to being tail recursive and then rewriting it iteratively is actually mechanical (there are compilers that can optimize tail recursion into a loop). The solution I wrote was not based on a tail recursive solution (I just worked through it and tried to remember how it worked; I am going to have to assign something like this in the next week so most of this is fresh in my mind).

-bcl
0

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