/* Pseudo code for unsorted linked list using dynamic memory
   Watch out for possible bugs!
 */
 

/********************************************************
 * insert node, assume node.next = null
 * return head (because head will change)
 ********************************************************/

if (head == null) head = node;
else {
   node.next = head;
   head = node;
   }
return head;
   
/********************************************************
 * delete single node, return reference to deleted node
 ********************************************************/

if (head == null) return null;
else {
   retValue = head;
   head = head.next;
   return retValue;
   }
   

/********************************************************
 * find node with value== key.  return reference to found node
 ********************************************************/
 
for (p = head; p != null && p.value != key; p = p.next) ;
return p;


/********************************************************
 * count number of nodes in list
 *   iterative solution
 ********************************************************/
 for (count = 0, p = head;
      p != null;
      count++, p = p.next) ;
 return count;
 
 /* more verbose iterative solution */
 count = 0;
 for (p = head; p!= null; p = p.next)
     count++;

 return count;
 
 /*  recursive solution  */
 int doCount(list) {
    if (list == null) return 0;
    return 1 + doCount(list.next);
    }
    

/********************************************************
 * delete (first) element with value == key
 ********************************************************/
 
 // empty list
if (head == null) return null;

// first element is desired element
if (head.value == key) {
   p = head;
   head = head.next;
   return p;
   }

// desired element is buried in list
// p points to candidate node, q is one node behind
p = head.next;
q = head;
while (p != null && p.value != key) {
   q = p;
   p = p.next;
   }

// p now points to desired element, update previous element
//  in order to remove p from list, then return p
q.next = p.next;
return p;
    }