Question

Since I don’t have a machine to test this I kindly need your help here.

• If I assign L.head = NULL will the L gets empty because there is
  no head ?

• If I assign L.head = L.next.next(3rd node) the previous two nodes
  will be as a Garbage Collector (assuming using Java) correct?

My attempt to write a method cutToInteger for the photo below is the following, correct it if I’m wrong:

void cutToInteger (IntSLList L , int n){

    IntSLList tmp =L.head ;
    while( tmp != NULL || !tmp.into.equals(n)){
            tmp=tmp.next;
    }
    L.head = tmp;
}

The implementation seems easy but the logic of the nodes becoming a garbage data to get removed always confuses me.

UPDATE: Here is the Question for the above screenshot

A method void cutToNumber(IntSLList L, int n) that cuts an integer singly linked list L starting from the head until it reaches integer n. If n is not in L, the list becomes empty.

Answer 1

1. in your loop it should be && operator rather than || ,otherwise it always will stop only at the end of the link.

2. n is int, so you have to use ==\!= operators for compare it with other int.

the method should be something like :

void cutToInteger (IntSLList L , int n){    
    IntSLList tmp =L.head;
    while( tmp != NULL && tmp.into != n ){
         tmp=tmp.next;
    }
    L.head = tmp;
}