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Advanced Data Structures. Chapter 16. Advanced Data Structures. Data structures that take control of organizing elements Elements not in fixed positions Advantage – better performance Adding Removing Finding. Set. Unordered collection of distinct elements Fundamental operations
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Advanced Data Structures Chapter 16
Advanced Data Structures • Data structures that take control of organizing elements • Elements not in fixed positions • Advantage – better performance • Adding • Removing • Finding
Set • Unordered collection of distinct elements • Fundamental operations • Adding an element • Removing an element • Containment testing (is something there) • Listing – arbitrary order • Duplicates • Aren’t permitted • Attempt to add – ignored • Attempt to remove non existent data - ignored
Data Structures for Sets Hash tables Trees Uses set interface
Hash Set • Create a Hash Set of Strings Set<String> names = new HashSet<String>(); • Adding and deleting names.add(“Romeo”); names.remove(“Juliet”); • Test to see if element is in set if (names.contains(“Juliet”);
Hash Set Iterate Iterator <String> iter=names.iterator(); while (iter.hasNext()) { String name = iter.next(); do something with the string ……… }
Hash Set For Each Loop for (String name: names) { do something with names { • Iterating through elements • Does not visit in order inserted • Set implementation rearranges for quick location • Cannot add in iterator position • Create a print method • Should be usable by has or tree
Maps • Data type • Associates keys and values • A function from one set(key) to another set(value) • Every key has unique value • A value may have multiple keys • Two kinds of implementation • Hash Map • Tree Map • Both implement Map Interface
Creating A Map Values Keys Romeo Adam Eve Juliet Green Red Pink Map<String,Color> favoriteColors = new HashMap<String, Color?();
Map • Add an association • favorite.put(“Juliet”,Color.PINK); • Change a value • favorite.put(“Juliet”,Color.RED); • Return the value • Color julietFavoriteColor = favorite.get(“Juliet”); • If you ask for a value with no key return a null • Remove key and value • favorite.remove(“Juliet”);
Map Find all keys and values in map Set<String> keySet = m.keySet(); For(String key: KeySet) { Color value=m.get(key); System.out.println(Key + “->” + value); }
Hash Table • Find elements in a data structure quickly • No linear search through all elements • Can implement sets and maps • Utilizes hash codes • Function that computes an integer value from an object • Integer value = has code • Object class has a hashCode method • Int h = x.hashCode(); • Collision – 2 objects with same hash code • Good has functions minimizes collision • Hash Code – index into hash table
Hash Table • Simplest form • Array index to hash table • In array insert each object @ location of its hash code • If anything already in array – object is present
Hash Table [70068] [74478] [74656]
Problem with this idea • Not possible to allocate an array that is large enough to hold all possible integer index positions • Must reduce hash function to fit the array size int h = x.hashCode(); if (h<0) h = -h; position = h.%buckets.length; • This approach increases likely hood of collisions
What to Do [65] Harry Sue Nina [66] Susannah [67] [68] Larry [69] [70] Adam [71] [72] Juliet Katherine Tony Create a hash table that is implemented as an array of buckets. Buckets are a sequence of nodes that hold elements with the same hash code.
Algorithm Finding Object in Hash Table • Compute hash code • Reduce to table size using modulo • Produces index into hash table int h = x.hashCode(); if (h<0) h = -h; position = h.%buckets.length; • Iterate through the elements of the bucket at position h • Check for match
Efficiency Depends on number of collisions or buckets Few buckets – quick O(1) Many buckets – much slower
Computing Hash Codes • From String int h = 0; for (int i=0; I < s.length(); i++) h = h+s.charAt(i); • What is the problem with this? • Same hash code for permutations of same letters (eat tea)
Standard Library Method final int HASH_MULTIPLIER = 31; Int h=0; for(int i=0;i<s.length();i++) h = HASH_MULTIPLIER*h+s.charAt(i); eat 31*(31*’e’ +’a’) + ‘t’ = 100184 31*(31*101+97) + 116 tea 31*(31*’t’ +’e’) + ‘a’ = 114704 31*(31*116+101) + 97
How to Create Hash Code for Your Object • Example Coin class public class Coin { final int HASH_MULTIPLIER = 29; public int hashCode() { int h1 = name.hashCode(); int h2 – new Double(value.hashCode(); int h = HASH_MULTIPLIER*h1+h2; return h; } • In a hash map only the keys are hashed.