Abstract Data Types (ADTs)
Array-based Stack
normal configuration wrapped-around configuration
We use the modulo operator (remainder of division) Operation enqueue throws an exception if the array is full This exception is implementation-dependent Operation dequeue throws an exception if the queue is empty This exception is specified in the queue ADT
p A A p p B q B p A C B A C A B B C C X q X D p D C A B C
Operation size, is. Empty at. Rank, rank. Of, elem. At. Rank first, last, before, after replace. Element, swap. Elements replace. At. Rank insert. At. Rank, remove. At. Rank insert. First, insert. Last insert. After, insert. Before remove Array List
The Tree Structure
Tree Terminology subtree
Tree ADT
The Tree Structure
Tree Terminology subtree
Tree ADT
Depth and Height
Preorder Traversal
Postorder Traversal
Binary Tree
Binary Tree Examples 2 3 a b 1 Want a fast meal? How about coffee? Starbucks Spike’s On expense account? Al Forno Café Paragon
Properties of Binary Trees
Binary. Tree ADT
Inorder Traversal
Print Arithmetic Expressions 2 3 a 1 b
2 3 5 1 2
Euler Tour Traversal 2 3 5 1 2
Template Method Pattern
Specializations of Euler. Tour
Data Structure for Trees B D A C F E
Data Structure for Binary Trees B A D C E
Java Implementation
Trees in JDSL
Priority Queue ADT
Total Order Relation
Comparator ADT
Sorting with a Priority Queue
Selection-Sort
Insertion-Sort
In-place Insertion-sort
What is a heap
Height of a Heap
Heaps and Priority Queues We can use a heap to implement a priority queue We store a (key, element) item at each internal node We keep track of the position of the last node For simplicity, we show only the keys in the pictures
Insertion into a Heap
Upheap
Removal from a Heap
Downheap
Updating the Last Node
Heap-Sort
Merging Two Heaps
Example
Example (contd. )
Example (contd. )
Example (end)
Analysis
Hash Functions
Hash Code Maps
Hash Code Maps (cont. )
Compression Maps
Collision Handling
Linear Probing
Search with Linear Probing
Updates with Linear Probing
Performance of Hashing
Universal Hashing
Proof of Universality (Part 1) Let f(k) = ak+b mod p Let g(k) = k mod N So h(k) = g(f(k)). f causes no collisions: So a(j-k) is a multiple of p But both are less than p So a(j-k) = 0. I. e. , j=k. (contradiction) Thus, f causes no collisions.
Proof of Universality (Part 2)
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