35->10->35->47->9->26->42->35->17->24->8-> Putting this linked list through the

Question

35->10->35->47->9->26->42->35->17->24->8->

Putting this linked list through the function below it takes the last half of the list and puts it in front. I understand that, but I dont understand the line of code that i commented on. what does (last->next = first) do in the function??

Void SLL::sj() {

SNode *tmp = first;

for(int i = 0; i < size/2; i++) {

tmp = tmp->next;

}

last->next = first //What does this do in the function?? does it set 8 equal to first, which would be 35's previous??

first = tmp->next;

tmp->next = NULL;

}

Explanation / Answer

Void SLL::sj() {

SNode *tmp = first;        // first is storing the pointer to the first element ie 35

for(int i = 0; i < size/2; i++) {            // size =11 ,size/2= 5, i=0 to 4, 5 elements are traversed, 10,35,47,9,26

tmp = tmp->next;                           // temp =26

}

//last is storing the pointer to the last element in list ie 8 .Now last->next = null and first =35

last->next = first // by making last->next =first , the element after 8 will be 35 in the list as first is 35

first = tmp->next;     // value in temp->next = 26->next = 42, so first element in the new list is 42

tmp->next = NULL;   // 26->next = null so 26 will be the last element in the list

}

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