21+-+Templates - 9/1/10 Linked Lists Double-ended list What...

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9/1/10 1 Templates EECS 280 Programming and Introductory Data Structures 1 Linked Lists Double-ended list 2 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. first Linked Lists Double-ended list 3 The new representational invariant 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. Linked Lists Double-ended list 4 So, in an empty list, both data members point to NULL. However, if the list is non-empty, they look like this: Note: Adding this new data member requires that every method (except isEmpty ) be re-written. In lecture, we'll only write insertLast . first last Linked Lists Double-ended list 5 First, we create the new node, and establish its invariants: void IntList::insertLast(int v) { node *np = new node; np->next = NULL; np->value = v; ... } Linked Lists Double-ended list 6 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 two broken invariants. The "old" last->next element (incorrectly) points to NULL, and the last field no longer points to the last element. first last np
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9/1/10 2 Linked Lists Double-ended list 7 void IntList::insertLast(int v) { node *np = new node; np->next = NULL; np->value = v; if (isEmpty()) { first = last = np; } else { last->next = np; last = np; } } first last np Linked Lists Double-ended list 8 This is efficient, but only for insertion. Question : Why is removal from the end expensive? first last np first last Linked Lists Double-ended list 9 To make removal from the end efficient, as well, we have to have a “doubly-linked” list, so we can go forward and backward. 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 stay 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. Linked Lists
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21+-+Templates - 9/1/10 Linked Lists Double-ended list What...

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