Question

1) Write a sequence of C++ statements to add a node to the front [1] of the list. Make sure you put 4 in the new Node.

2) Write a while loop here that goes through the list until P points to the node right before the rear node. [2]

P = Front; // to start with

3) Write a for loop here that makes P stop at the (J-1)th node. [2] Think how many times the pointer needs to move.

P = Front; // to start with

Answer 1

1. Let,

front is the pointer pointing to the first node (front) of the list.

A Node class has two fields namely data (which contains the data/value ) and next (which is a pointer to the next node).

Node * temp = NULL;

Node temp = new Node()

temp->data= 4

temp->next = front

front = temp

2. P = front;

while (p->next->next != NULL){ //p->next->next always lloks at the next value of one node after the current one

p = p->next;

}

3. Suppose j = 4. We want pointer P to point to j-1 i.e 3rd node in the list. Initially P is already pointing to the first (front node). To come to 3rd node, we have to move P two times. On first move, P comes to 2nd node, on second move it comes to 3rd node. So , in general, the loop has to execute j-2 times.

P = front;

for(int i=1; i<=j-2; i++) {

p = p->next;   

}

Please do let me know if you have any questions/doubts.