8.1 KiB
8.1 KiB
Lecture 10 --- Iterator Implementation & Linked Lists
- Iterator Implementation (in Vectors)
- Building our own basic linked lists:
– Stepping through a list
– Basic operations
10.1 Review: Iterators and Iterator Operations
What is the output of this program?
#include <list>
#include <iostream>
int main(){
std::list<int> lst;
lst.push_back(150);
lst.push_back(250);
lst.push_back(350);
lst.push_back(450);
std::list<int>::iterator itr;
itr = lst.begin();
++itr;
*itr += 5;
std::list<int>::iterator itr2 = lst.begin();
while(itr2 != lst.end()){
std::cout << *itr2 << std::endl;
itr2++;
}
}
10.2 Iterator Implementation
Here we extend our implemention of vector so as to include iterators. Review the written code here: vec.h and vec_main.cpp.
10.3 Working towards our own version of the STL list
- Our discussion of how the STL list<T> is implemented has been intuitive: it is a “chain” of objects.
- Now we will study the underlying mechanism — linked lists.
- This will allow us to build custom classes that mimic the STL list class, and add extensions and new features.
10.4 Objects with Pointers, Linking Objects Together
- The two fundamental mechanisms of linked lists are:
– creating objects with pointers as one of the member variables, and
– making these pointers point to other objects of the same type. - These mechanisms are illustrated in the following program:
template <class T>
class Node {
public:
T value;
Node* ptr;
};
int main() {
Node<int>* ll; // ll is a pointer to a (non-existent) Node
ll = new Node<int>; // Create a Node and assign its memory address to ll
ll->value = 6; // This is the same as (*ll).value = 6;
ll->ptr = NULL; // NULL == 0, which indicates a "null" pointer
Node<int>* q = new Node<int>;
q->value = 8;
q->ptr = NULL;
// set ll's ptr member variable to
// point to the same thing as variable q
ll->ptr = q;
cout << "1st value: " << ll->value << "\n"
<< "2nd value: " << ll->ptr->value << endl;
}
- Play this animation to see how this program works.
10.5 Definition: A Linked List
- The definition is recursive: A linked list is either:
- Empty, or
- Contains a node storing a value and a pointer to a linked list.
- The first node in the linked list is called the head node and the pointer to this node is called the head pointer. The pointer’s value will be stored in a variable called head.
10.6 Visualizing Linked Lists
- The head pointer variable is drawn with its own box. It is an individual variable. It is important to have a separate pointer to the first node, since the “first” node may change.
- The objects (nodes) that have been dynamically allocated and stored in the linked lists are shown as boxes,
with arrows drawn to represent pointers.
- Note that this is a conceptual view only. The memory locations could be anywhere, and the actual values of the memory addresses aren’t usually meaningful.
- The last node MUST have NULL for its pointer value — you will have all sorts of trouble if you don’t ensure this!
- You should make a habit of drawing pictures of linked lists to figure out how to do the operations.
10.7 Basic Mechanisms: Stepping Through the List
- We’d like to write a function to determine if a particular value, stored in x, is also in the list.
- We can access the entire contents of the list, one step at a time, by starting just from the head pointer. – We will need a separate, local pointer variable to point to nodes in the list as we access them. – We will need a loop to step through the linked list (using the pointer variable) and a check on each value
10.8 Exercise: Write is_there
template <class T> bool is_there(Node<T>* head, const T& x) {
}
- If the input linked list chain contains n elements, what is the order notation of is_there?
10.9 Overview: Adding an Element at the Front of the List
- Goal: place a new node at the beginning of the list.
- We must create a new node.
- We must permanently update the head pointer variable's value. Therefore, we must pass the pointer variable by reference.
10.10 Exercise: Write push_front
template <class T> void push_front( Node<T>* & head, T const& value ) {
}
-
Play this animation to see how push_front works.
-
If the input linked list chain contains n elements, what is the order notation of the implementation of push_front?
10.11 Basic Mechanisms: Pushing on the Back
- Goal: place a new node at the end of the list.
- We must step to the end of the linked list, remembering the pointer to the last node.
- This is an O(n) operation and is a major drawback to the ordinary linked-list data structure we are discussing now. We will correct this drawback by creating a slightly more complicated linking structure in our next lecture.
- We must create a new node and attach it to the end.
- We must remember to update the head pointer variable’s value if the linked list is initially empty. – Hence, in writing the function, we must pass the pointer variable** by reference.**
10.12 Exercise: Write push_back
template <class T> void push_back( Node<T>* & head, T const& value ) {
}
-
Play this animation to see how push_back works.
-
If the input linked list chain contains n elements, what is the order notation of this implementation of push_back?
10.13 Inserting a Node into a Singly-Linked List
- With a singly-linked list, we’ll need a pointer to the node before the spot where we wish to insert the new item.
- If p is a pointer to this node, and x holds the value to be inserted, then the following code will do the insertion. Draw a picture to illustrate what is happening.
Node<T> * q = new Node<T>; // create a new node
q -> value = x; // store x in this node
q -> next = p -> next; // make its successor be the current successor of p
p -> next = q; // make p's successor be this new node
-
Play this animation to see how this code snippet works.
-
Note: This code will not work if you want to insert x in a new node at the front of the linked list. Why not?
10.14 Removing a Node from a Singly-Linked List
- The remove operation itself requires a pointer to the node before the node to be removed.
- Suppose p points to a node that should be removed from a linked list, q points to the node before p, and head points to the first node in the linked list. Note: Removing the first node is an important special case.
- Write code to remove p, making sure that if p points to the first node that head points to what was the second node and now is the first after p is removed. Draw a picture of each scenario.
10.15 Exercise: Singly-Linked List Remove All
Write a recursive function to delete all nodes in a linked list. Here’s the function prototype:
template <class T> void RemoveAll(Node<T>*& head) {
}
10.16 Exercise: Singly-Linked List Copy
Write a recursive function to copy all nodes in a linked list to form an new linked list of nodes with identical structure and values. Here’s the function prototype:
template <class T> void CopyAll(Node<T>* old_head, Node<T>*& new_head) {
}