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nietod...Help with algorithm

Posted on 2000-02-14
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You solved the josephus algorithm i posted. I'm trying to implement it now with a circular linked list. An have no idea where to start it. this is the class im using:

// Circular Linked List Class
#ifndef CIRCLIST_H
#define CIRCLIST_H

#include <iostream.h>

template <class T>
class CircularList {
public:
   CircularList();
   ~CircularList();
   int Add(T item);                     // return -1 if out of memory
   int Remove(int index);
   int Get(int index, T&item) const;    // return -1 if not gotten
   int Find(T item) const;              // return -1 if not found
   int Empty() const { return size==0; }
   int Size() const { return size; }
   struct Node {
      T data;
      Node *left, *right;
   };
private:
   Node *head;
   int size;
};


//----------------------------//
template <class T>
CircularList<T>::CircularList() {
   head=new Node;
   head->right=head;
   head->left=head;
   size=0;
}


//----------------------------//
template <class T>
CircularList<T>::~CircularList() {
   while (Size()>0) Remove(0);
   delete head;
}


//----------------------------//
template <class T>
int CircularList<T>::Add(T item) {
   Node *temp, *cur;
   temp = new Node;                    // get a new node and fill with data
   if (temp==NULL) { cout << "Out of memory for linked list node allocation: Add() failed." << endl; return -1; }
   temp->data = item;
   cur = head->right;

   while(cur != head  &&  cur->data < temp->data) {
      cur=cur->right;
   }

   temp->right = cur;
   temp->left = cur->left;
   cur->left->right = temp;
   cur->left = temp;

   size++;                             // adjust size
   return 1;
}



//----------------------------//
template <class T>
int CircularList<T>::Remove(int index) {
   if (Empty() || index<0 || index>=Size() ) return -1;
   Node *cur;
   cur = head->right;
   for (int i=0; i<index; i++) {
      cur=cur->right;
   }                                   // at end of loop, cur is pointing
                                       // to the node to delete
   cur->left->right = cur->right;
   cur->right->left = cur->left;

   delete cur;                         // free up space
   size--;                             // adjust size
   return size;
}


//----------------------------//
template <class T>
int CircularList<T>::Get(int index, T &item) const {     // similar to Remove
   if (Empty() || index<0 || index>=Size() ) return -1;
   Node *cur;
   cur = head->right;
   for (int i=0; i<index; i++) {
      cur=cur->right;
   }                                   // at end of loop, cur is pointing
                                       // to the node to get
   item=cur->data;                     // transfer data
   return 1;
}


//----------------------------//
template <class T>
int CircularList<T>::Find(T item) const {
   if (Empty()) return -1;
   int index=0;
   Node *cur=head->right;
   while (cur!=head && item!=cur->data) {
      cur=cur->right;
      index++;
   }
   return (cur==head ? -1 : index);
}

#endif



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Question by:milalik
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LVL 22

Expert Comment

by:nietod
ID: 2523033
What exactly is the question?

I assume the josephus algorithm needs to store data (I'm not familar with the algorithm) so the data will be stored in this circular queue.
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Accepted Solution

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nietod earned 150 total points
ID: 2523172
I found the algorithm from your other question.

Since this is probably accademic, I can't give you the answer--that is unethical--but I'll give you some help to get you started.  Ask if you have questions or need more help.

Start with all the items in the queue  Then use a loop to remove items from the queue.  Each item removed is placed back in the queue until you come to an item that should be "skipped".  If the item should be skipped don't place it back in.  When you come to a case where you remove an item from the queue and the queue is left empty (i.e. there had been 1 item in the queue), then you are done.  The item you just removed is te last one.
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Author Comment

by:milalik
ID: 2523442
Okay so I add the items to the list in the main. Then perform the Remove with the conditions every m-time..and when the only one number ends I'm done?
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LVL 22

Expert Comment

by:nietod
ID: 2523463
That's it.
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