lecture19 - Last time: * Linked structures * single-ended...

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Last time: * Linked structures * single-ended list Today: * Double-ended list * doubly-linked list * The "container-of-references" model * Intro to templates +++++++++++++++++++++++++++++++++++++++++++++++++++++++++ What if we wanted to insert something at the end of the list? Intuitively, with the current representation, we'd need to walk down the list until we found "the last element", and then insert it there. That's not very efficient, because we'd have to examine every element to insert anything at the tail. Instead, we'll change our concrete representation to track both the front and the back of our list. The new rep has *two* node pointers: class IntList { node *first; node *last; public: ... }; The invariant on first is unchanged. The invariant on "last" is: last points to the last node of the list if it is not empty, and is NULL otherwise. So, in an empty list, both data members point to NULL. However, if the list is non-empty, they look like this: +---+ +---+ +---+ +---+ first----> | -----> | -----> | -----> | -----\ +---+ +---+ +---+ +---+ \ ^ --- last-----------------------------------/ - Note: adding this new data member requires that *every* method (except isEmpty) be re-written. In lecture, we'll only re-write insertLast. First, we create the new node, and establish its invariants: void IntList::insertLast(int v) { node *np = new node;
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np->value = v; ... } To actually insert, there are two cases---if the list is empty, we need to reestablish the invariants on first *and* last---the new node is both the first and last node of the list. If the list is *not* empty, there are still two broken invariants. The "old" last element (incorrectly) points to NULL, and the "last" field no longer points to the last element. void IntList::insertLast(int v) { node *np = new node; np->next = NULL; np->value = v; if (isEmpty()) { first = np; } else { last->next = np; } last = np; } Now how many pointers must be examined? +++++++++++++++++++++++++++++++++++++++++++++++++++++++++ This is efficient, but only for insertion. To make removal from the end efficient, as well, we have to have a doubly-linked list, so we can go forward *and* backward from any inidivdual node. To do this, we're going to change the representation yet again. In our new representation, a node is: struct node { node *next; node *prev; int value; } The next and value fields are the same. The "prev" field's invariant is: The "prev" field points to the previous node in the list, or NULL if no such node exists. With this representation, an empty list is unchnaged
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lecture19 - Last time: * Linked structures * single-ended...

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