c# - Linked list addition of two minimum at last node like tree -

i have switched c# c++. under situation using linked list contains nodes follows : (which sorted in increasing order)

1->1->1->44->46->48->49->50->null 

what have :

(1)add first 2 nodes. (2)put obtained @ last node. so, here after first addition llist :(1+1=2)

1->1->1->44->46->48->49->50->2->null 

(3)again add 2 minimum nodes , put sum @ last node(but don't add nodes added).

(4) time 2 minimum 1 , 2 because "1" , "1"(the first two) added. llist becomes :

 1->1->1->44->46->48->49->50->2->3->null 

(5) 2 minimum 3 , 44 ,so

1->1->1->44->46->48->49->50->2->3->47->null 

(6) 2 minimum 46 , 47(see @ last have 47 second smallest) , 1->1->1->44->46->48->49->50->2->3->47->93->null (7) 48 , 49,

1->1->1->44->46->48->49->50->2->3->47->93->97->null 

(8) next minimums "50" , "93" left so,

1->1->1->44->46->48->49->50->2->3->47->93->97->143->null 

(9) finally: 97 , 143 left, so

 1->1->1->44->46->48->49->50->2->3->47->93->97->143->240->null 

there 1 element left(240) stop here.

could 1 please me in making it's algorithm ?

my idea: implement algorithm : here "freq" contains value , left , right tree left , right.

    while (front != rear)      {          if (counter == 0)          {               console.writeline("check1");             temp = new node();             temp.freq = front.freq + front.next.freq;             front.is_processed = 1;             front.next.is_processed = 1;             temp.is_processed = 0;             temp.left = front;             temp.right = front.next;             temp.next = null;             rear.next = temp;             front = front.next.next;             rear = rear.next;             remaining = count_remaining();             if (remaining == 1)              {                 break;             }         }           if (rear.freq.equals(front.freq))          {              console.writeline("check2");             temp = new node();             temp.freq = front.freq + rear.freq;             rear.is_processed = 1;             front.is_processed = 1;             temp.is_processed = 0;             temp.left = front;             temp.right = rear;             temp.next = null;             rear.next = temp;             rear = rear.next;             remaining = count_remaining();             if (remaining == 1)              {                 break;             }         }         if (rear.freq < front.freq)          {            console.writeline("check3");             node pmin1 = null;             node pmin2 = null;             front_rear(ref pmin1, ref pmin2);             temp = new node();             temp.freq = pmin1.freq + pmin2.freq;             pmin1.is_processed = 1;             pmin2.is_processed = 1;             temp.is_processed = 0;             temp.left = pmin2;             temp.right = pmin1;             temp.next = null;             rear.next = temp;              rear = rear.next;             remaining = count_remaining();             if (remaining == 1)              {                 break;             }                            }             if (rear.freq > front.freq)         {                console.writeline("check4");             node pmin1 = null;             node pmin2 = null;             front_rear(ref pmin1, ref pmin2);             temp = new node();             temp.freq = pmin1.freq + pmin2.freq;             pmin1.is_processed = 1;             pmin2.is_processed = 1;             temp.is_processed = 0;             temp.left = pmin2;             temp.right = pmin1;             temp.next = null;             rear.next = temp;             rear = rear.next;             remaining = count_remaining();             if (remaining == 1)             {                 break;             }         }                     counter++;     }  

but surprisingly prints : (it not @ go "check3" if (rear.freq < front.freq) , can see step5 above @ condition don't print check3 )

check1 check4 check4 check4 check4 check4 check4 

why going in condition if (rear.freq > front.freq) ? (it's actualy tree using linked list) parent node sum of 2 minimum nodes.

your code sloppy, i'll give basic framework use

node first = findlowestunprocessednode(); node second = findlowestunprocessednode();  while(first != null && second != null) {      node sumnode = new node(first, second)      addtoendoflist(sumnode);        node first = findlowestunprocessednode();     node second = findlowestunprocessednode(); } 

here functions make yourself. these shouldn't hard

• findlowestunprocessednode(...): finds lowest unprocessed node , marks processed. if nodes processed, returns null. might find want add parameters method.

• node(node first, node second): add constructor node class. should initialize node have first left , second right, , it's freq should sum of first.freq , second.freq.

• addtoendoflist(node newnode);: adds node end of list.

please note these abstract place-holder methods can , should add parameters them, or make them instance methods if think out.