25 march 2013birkbeck college, u. london1 introduction to programming lecturer: steve maybank...

25 March 2013 Birkbeck College, U. London 1

Introduction to Programming

Lecturer: Steve Maybank

Department of Computer Science and Information Systems

sjmaybank@dcs.bbk.ac.ukSpring 2013

Week 11: Examples of Algorithms

JavaLab 9, Ex. 1 (1)

Write a method

public static int[] reverseArray(int[] data)

such that reverseArray returns the reverse of the array data. Call reverseArray from main.

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JavaLab 9, Ex. 1 (2)import java.util.Arrays;/** * This program reverses the order of the elements in an

array * @author S.J. Maybank * @version 25 March 2013 */public class ReverseArray{ // main defined here // reverseArray defined here}

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JavaLab 9, Ex. 1 (3)public static void main(String[] args){ int[] data1 = {1, 2, 3, 4, 5}; int[] data2 = {}; int[] data3 = {1, 2, 3, 4}; int[] data1R = reverseArray(data1); int[] data2R = reverseArray(data2); int[] data3R = reverseArray(data3); System.out.println("Original array: "+Arrays.toString(data1)); System.out.println("Reversed array:

"+Arrays.toString(data1R)); // data2, dataR and data3, data3R printed out similarly}

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JavaLab 9, Ex. 1 (4)public static int[] reverseArray(int[] data){ int[] dataReversed = new int[data.length]; for(int i = 0; i < data.length; i++) { dataReversed[data.length-1-i] = data[i]; } return dataReversed;}

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JavaLab 9, Ex. 2 (1)/** * A program to apply certain methods to an array of integers. * @author S.J. Maybank * @version 25 March 2013 */public class ArrayMethods{

// main // printArray

// productElements// numberNegativeElements

}

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JavaLab 9, Ex. 2 (2)

public static void main(String[] args){ int[] data = {1, 2, 3, -1, -3}; printArray(data); System.out.println("Product elements:

"+productElements(data)); System.out.print("Number of elements strictly less than 0:

"); System.out.println(""+numberNegativeElements(data));}

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JavaLab 8, Ex. 2 (3)public static void printArray(int[] data){ for(int i = 0; i < data.length; i++) { System.out.print(data[i]); if (i < data.length-1) { System.out.print(" "); } } System.out.println();}

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JavaLab 9, Ex. 2 (4) public static int productElements(int[] data)

{ int product = 1; for(int i = 0; i < data.length; i++) { product *= data[i]; } return product; }

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JavaLab 9. Ex. 2 (5)public static int numberNegativeElements(int[] data){ int n = 0; for(int i = 0; i < data.length; i++) { if(data[i] < 0) { ++n; } } return n;}

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Overview

Test to see if two appointments overlap (JFE, R3.11)

Linear search of an array (JFE, Section 6.3.5)

Binary search of an array (JFE, end of Section 6.3)

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Appointments

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timea1 a2 a3 a4

Non-overlapping appointments: [a1, a2] and [a3, a4]Overlapping appointments:

[a1, a3] and [a2, a4][a1, a4] and [a2, a3]

How to decide when two appointments overlap?

Overlapping Appointments Let the appointments be [s1, e1] and [s2, e2].

Let s be the latest start and let e be the earliest end.

If s > e, then one appointment begins after the other has finished.

Conversely, let t be any time such that s ≤ t ≤ e. The time t is in [s1, e1] because

s1 s ≤ t ≤ e e1 The time t is in [s2, e2] because

s2 s ≤ t ≤ e e2

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Pseudo Code for overLappingAppointments

Inputs: times s1, e1 and s2, e2 for two appointments.

Output: true if the appointments overlap and false otherwise

if (s1 > s2)s = s1 else s = s2

if (e1 < e2)e = e1 else e = e2

if (s < e) return true else return false

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The Method overLappingAppointments

public static boolean overLappingAppointments(int s1, int e1, int s2, int e2)

{int s, e;if(s1 > s2){s = s1;} else {s = s2;}if(e1 < e2){e = e1;} else {e = e2;}return s < e;

}

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Inputs and Output for Linear Search

Inputs: 1D integer array data and an integer e.

Output: if data contains e then an integer pos such that

data[pos] == e,otherwise –1.

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Pseudo Code for Linear Search

1. Step through the valid indices pos for the array data,

if data[pos] == e, then return pos.

2. If all the valid indices have been checked without finding e, then return -1

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The Method linearSearchpublic static int linearSearch(int[] data, int e){

int pos = 0;while (pos < data.length){

if(data[pos] == e){return pos;}++pos;

}return –1;

}

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The Method linearSearch2public static int linearSearch2(int[] data, int e){

int pos = data.length-1;while (pos>=0){

if(data[pos] == e){return pos;}pos--;

}return –1;

}

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Inputs and Output for Binary Search

Inputs: 1D sorted integer array data and an integer e.

Output: if data contains e then an integer pos such that

data[pos] == e,otherwise –1.

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Strategy for Binary Search

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1 3 7 8 11 12 20

low high

Mark out a section of the array data using indices low, highFind an index i between low and highCompare data[i] with e and update low, high accordingly

i

Pseudo Code for Binary Search

Set low equal to the least index for data.Set high equal to the largest index for data.while (low <= high){

Find an index pos between low and high.if (data[pos] == e) then return pos.if (data[pos] < e) then high = pos-1.if (data[pos] > e) then low = pos+1.

}return –1.

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The Method binarySearchpublic static int binarySearch(int[] data, int e){

int low = 0, high = data.length-1, pos = 0;while(low <= high){

pos = (low+high)/2;if (data[pos] == e){return pos;}if (data[pos] < e){low = pos+1;} // look in second halfelse{high = pos-1;} // look in first half

}return –1;

}

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Example of Binary Search Inputs: data = {1, 3, 5, 7, 9, 11}, e = 6.low = 0, high = 5, pos = 2.data[pos] < e, thus low = pos+1 = 3.low = 3, high = 5, pos = 4.data[pos] > e, thus high = pos-1 = 3.low = 3, high = 3, pos = 3.data[pos] > e, thus high = pos-1 = 2.low = 3, high =2(low <= high) == false, search terminates.

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