www.hope.ac.uk faculty of sciences and social sciences hope data structures 2 stewart blakeway...

www.hope.ac.uk Faculty of Sciences and Social Sciences

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Data Structures 2

Stewart Blakeway

[email protected]


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Aims of the Presentation

• To demonstrate Arrays in Java– Defining the Array object– Determining the Range– Assigning a value to an element– Sorting the Array– Searching the Array


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Quick question – What is Shopping[1] and Shopping[4] ?

• Shopping[1] is Crisps• Shopping[4] does not exist

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0 Wine1 Crisps2 Nuts3 Bread

Shopping


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Another question


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Final question


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What we have done before in data structures

• A record is a collection of different data types

• An array is a fixed size collection of same data types– Single-dimension– Multi-dimension

• A sequence is a variable size collection of same data types– Stack– Queue

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Operations on arrays: assignment

• Scores[4] = 29;

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Operations on arrays: retrieval

• What would be the output of:

System.out.println(Scores[3]+Scores[10]);

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Example of retrieval

• Write a fragment of Java that would display the third name in the array.

System.out.println(PlayOffTeams[3]);

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• Write a fragment of Java that would display all the names in the array.

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int i;for (i = 1; i<=4; i++) { System.out.println(PlayOffTeams[i]); }

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Initialisation

• When arrays are created they should be initialised, otherwise they will possibly contain random numbers or NULL values

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The array Scores would want to start out life like this – each element contains 0


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Initialisation

• We can initialise Scores in Java like this:

int j;for (j=1; j<=10; j++) { Scores[j] = 0; }

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Searching and sorting arrays

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Searching an array

• Example: – How does a “hole in the wall” find your account

from your credit card details ?

• Need to search through the array of bank records

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Different kinds of search available

• Linear Search• Binary search

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Linear Search• Start at the beginning and examine each in

turn until a match found or end of records reached

– Fastest find in - • 1 comparison

– Slowest find in -• 1,000 comparisons

– Average find in - • 500 comparisons

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for simplicity, we will assume that each search takes 1ms


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• Search for a national insurance number

• Is a linear search appropriate ?

• No !

• A better solution in this case is a binary search

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Linear Search

Approximately 59 million people in the UK = Average Search Time of: 48.77 weeksMaximum Search Time of: 1y 43 weeks


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• Think of a number between 1 and 1000

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Binary Search


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• Think of a number between 1 and 1000• I can get that number in 10 yes/no questions

– is it less than 500? No– is it less than 750? No– is it less than 825? Yes– ... keep going

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Binary Search


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• Search for Evans in the array of names which have been sorted into alphabetical order

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Binary Search


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• 1 Ball• 2 Davies• 3 Evans• 4 Galt• 5 Hurst• 6 Martin• 7 Mason• 8 Moore• 9 Perkins• 10 Stephens

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See pages 37, 38 in booklet

Searching for Evans


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Searching for Evans


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Searching for Evans


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Searching for Evans


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Searching for Evans


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Searching for Evans


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Found Evans !

Searching for Evans


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Sorting

• We need to sort to– be able to use the fast binary search which depends

on the array being in ascending or descending order– produce reports in required order

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Sorting

• Many different ways of carrying out a sort on an array.

• We shall examine EXCHANGE SORT (Also known as the BUBBLE SORT)

• We shall look at others

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Exchange Sort

• Simple• Slow

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• 1 Davies• 2 Martin• 3 Perkins• 4 Evans• 5 Mason• 6 Ball• 7 Stephens• 8 Moore• 9 Hurst• 10 Galt

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No swap


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• 1 Davies• 2 Martin• 3 Perkins• 4 Evans• 5 Mason• 6 Ball• 7 Stephens• 8 Moore• 9 Hurst• 10 Galt

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No swap


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• 1 Davies• 2 Martin

• 3 Perkins• 4 Evans• 5 Mason• 6 Ball• 7 Stephens• 8 Moore• 9 Hurst• 10 Galt

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Swap


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• 1 Davies• 2 Martin• 3 Evans

• 4 Perkins• 5 Mason• 6 Ball• 7 Stephens• 8 Moore• 9 Hurst• 10 Galt

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Swap


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• 1 Davies• 2 Martin• 3 Evans• 4 Mason

• 5 Perkins• 6 Ball• 7 Stephens• 8 Moore• 9 Hurst• 10 Galt

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Swap


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• 1 Davies• 2 Martin• 3 Evans• 4 Mason• 5 Ball• 6 Perkins• 7 Stephens• 8 Moore• 9 Hurst• 10 Galt

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No Swap


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• 1 Davies• 2 Martin• 3 Evans• 4 Mason• 5 Ball• 6 Perkins

• 7 Stephens• 8 Moore• 9 Hurst• 10 Galt

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Swap


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• 1 Davies• 2 Martin• 3 Evans• 4 Mason• 5 Ball• 6 Perkins• 7 Moore

• 8 Stephens• 9 Hurst• 10 Galt

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Swap


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• 1 Davies• 2 Martin• 3 Evans• 4 Mason• 5 Ball• 6 Perkins• 7 Moore• 8 Hurst

• 9 Stephens• 10 Galt

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Swap


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• 1 Davies• 2 Martin• 3 Evans• 4 Mason• 5 Ball• 6 Perkins• 7 Moore• 8 Hurst• 9 Galt

• 10 Stephens

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At the end of this first pass,

Stephens is in the correct position alphabetically.

This is repeated for a second pass starting at position 1 and again and again until the array is

completely sorted.


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Java Exchange Sort

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Consider an array of 10,000 surnames, called names, indexed from 0 to 9,999.

for (i=1; i<=9999; i++)

{

for (j=0; j< 9999; j++)

{

if (names[j] > names[j+1])

{

temp = names[j];

names[j] = names[j+1];

names[j+1] = temp;

}

}

}


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Summary

• Operations on arrays– Initialisation– Assignment– Retrieval– Searching – linear and binary– Sorting – exchange sort

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