The diagrams in the following table represent a progression of inserts into an e
ID: 3807365 • Letter: T
Question
The diagrams in the following table represent a progression of inserts into an empty linked list. Match the following lines of code with the appropriate illustration on the left. Use each line only once: I. curr Ptr-head; 2. newNode ptr new ListElement (B,NULL) 3, currPtr- next-newNodePtra 4, head-newNodePtr 5, newNode Ptr new LiatElement (2,NULL) 6, currPtr new NodePtri Note what changes from one picture to the next.This represents the sequence of operations. 704F0 704F 704F0 704F0 704F NULL NULLExplanation / Answer
First Diagram : 5. newNodePtr = new ListElement(2, NULL);
Second Diagram :
5. newNodePtr = new ListElement(2, NULL);
4. head = newNodePtr;
Third Diagram :
5. newNodePtr = new ListElement(2, NULL);
4. head = newNodePtr;
3. currPtr = head;
Fourth Diagram:
5. newNodePtr = new ListElement(2, NULL);
3. currPtr = head;
4. head = newNodePtr;
2. newNodePtr = new ListElement(8, NULL);
Fifth Diagram:
5. newNodePtr = new ListElement(2, NULL);
3. currPtr = head;
4. head = newNodePtr;
2. newNodePtr = new ListElement(8, NULL);
3. currPtr->next = newNodePtr;
Sixth Diagram:
5. newNodePtr = new ListElement(2, NULL);
3. currPtr = head;
4. head = newNodePtr;
2. newNodePtr = new ListElement(8, NULL);
3. currPtr->next = newNodePtr;
6. currPtr = newNodePtr;