25 March 2013Birkbeck College, U. London1 Introduction to Programming Lecturer: Steve Maybank Department of Computer Science and Information Systems Spring 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. 25 March 2013Birkbeck College, U. London2

JavaLab 9, Ex. 1 (2) import java.util.Arrays; /** * This program reverses the order of the elements in an array S.J. Maybank 25 March 2013 */ public class ReverseArray { // main defined here // reverseArray defined here } 25 March 2013Birkbeck College, U. London3

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 } 25 March 2013Birkbeck College, U. London4

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; } 25 March 2013Birkbeck College, U. London5

JavaLab 9, Ex. 2 (1) /** * A program to apply certain methods to an array of integers. S.J. Maybank 25 March 2013 */ public class ArrayMethods { // main // printArray // productElements // numberNegativeElements } 25 March 2013Birkbeck College, U. London6

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)); } 25 March 2013Birkbeck College, U. London7

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(); } 25 March 2013Birkbeck College, U. London8

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; } 25 March 2013Birkbeck College, U. London9

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; } 25 March 2013Birkbeck College, U. London10

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) 25 March 2013Birkbeck College, U. London11

Appointments 25 March 2013Birkbeck College, U. London12 time a1a2a3a4 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 25 March 2013Birkbeck College, U. London13

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 25 March 2013Birkbeck College, U. London14

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; } 25 March 2013Birkbeck College, U. London15

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. 25 March 2013Birkbeck College, U. London16

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 March 2013Birkbeck College, U. London17

The Method linearSearch public static int linearSearch(int[] data, int e) { int pos = 0; while (pos < data.length) { if(data[pos] == e){return pos;} ++pos; } return –1; } 25 March 2013Birkbeck College, U. London18

The Method linearSearch2 public static int linearSearch2(int[] data, int e) { int pos = data.length-1; while (pos>=0) { if(data[pos] == e){return pos;} pos--; } return –1; } 25 March 2013Birkbeck College, U. London19

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. 25 March 2013Birkbeck College, U. London20

Strategy for Binary Search 25 March 2013Birkbeck College, U. London low high Mark out a section of the array data using indices low, high Find an index i between low and high Compare 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. 25 March 2013Birkbeck College, U. London22

The Method binarySearch public 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 half else {high = pos-1;} // look in first half } return –1; } 25 March 2013Birkbeck College, U. London23

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. 25 March 2013Birkbeck College, U. London24