Programming Languages and Techniques (CIS120e)

Transcription

1 Programming Languages and Techniques (CIS120e) Lecture 28 Nov. 15, 2010 Random- Access Data II

2 Programming with Arrays 2

3 Gocha! One of the most common difficulnes when manipulanng numerical indices into arrays is off- by- one errors 3

4 Fencepost Errors If you build a straight fence 100m long with posts 10m apart, how many posts do you need? static double[] subarray (double[] a, int m, int n) { double[] r = new double[n-m]; for (int i = m; i <= n; i++) r[i-m] = a[i]; return r; 4

5 Fencepost Errors Fencepost errors come from counting things rather than the spaces between them, or vice versa, or by neglecting to consider whether one should count one or both ends of a row. from the Jargon File 5

6 Avoiding Off- By- One Errors Use for- each loops (rather than low- level for loops) wherever possible Always think in terms of 0- based indices eliminate the term first element from your mental vocabulary instead think of element with index 0 Use first / upto convennons, not first / last when dealing with ranges of indices into an array, think of the starnng index as the first element in the range and the ending index as the first element not in the range Write tests that exercise your code at boundaries first element, last element, one- element array, empty array,... 6

7 Search 7

8 SpecificaNon What would be a good set of unit tests for this method? public static int search (int[] a, int key) 8

9 public void search_tests() { // Basic search int[] a = {1,3,5,7,9; asserttrue (linearsearch(a, 1)); asserttrue (linearsearch(a, 9)); asserttrue (linearsearch(a, 3)); assertfalse(linearsearch(a, 4)); // Empty array int[] a2 = {; assertfalse(linearsearch(a2, 4)); // Singleton array int[] a3 = {1; asserttrue(linearsearch(a3, 1)); assertfalse(linearsearch(a3, 4)); 9

10 ImplementaNon: Linear Search public static boolean linearsearch(int[] a, int key) { for (int i = 0; i < a.length; i++) { if (key == a[i]) return true; return false; How many comparisons does it take to find a given element, on average? 10

11 Binary Search If we can assume the array is sorted, we can search much faster!

12 Binary Search First try: Let s use the first / last convennon, just for fun... public static boolean binarysearch1 (int[] sorted, int key) { int first = 0; int last = sorted.length-1; while (first < last) { int mid = (first + last) / 2; // Compute midpoint if (key < sorted[mid]) { last = mid-1; // keep looking in bottom half else if (key > sorted[mid]) { first = mid + 1; // keep looking in bottom half else { return true; // Found it! return false; // Not found 12

13 Binary Search public static boolean binarysearch1 (int[] sorted, int key) { int first = 0; int last = sorted.length-1; while (first < last) { int mid = (first + last) / 2; // Compute midpoint if (key < sorted[mid]) { last = mid-1; // keep looking in bottom half else if (key > sorted[mid]) { first = mid + 1; // keep looking in bottom half else { return true; // Found it! return false; // Not found Example: search for 1 in {1,3,5,7,9. 13

14 Binary Search Second try... public static boolean binarysearch1 (int[] sorted, int key) { int first = 0; int last = sorted.length-1; while (first <= last) { int mid = (first + last) / 2; // Compute midpoint if (key < sorted[mid]) { last = mid-1; // keep looking in bottom half else if (key > sorted[mid]) { first = mid + 1; // keep looking in bottom half else { return true; // Found it! return false; // Not found 14

15 Binary Search Third try, using the first/upto convennon instead... public static boolean binarysearch3 (int[] sorted, int key) { int first = 0; int upto = sorted.length; while (first < upto) { int mid = (first + upto) / 2; // Compute midpoint if (key < sorted[mid]) { upto = mid; // keep looking in bottom half else if (key > sorted[mid]) { first = mid + 1; // keep looking in bottom half else { return true; // Found it! return false; // Not found 15

16 SorNng 16

17 SpecificaNon What would be a good set of unit tests for this method? public static void sort (int[] a) 17

18 SpecificaNon A preiy good public void sort_tests() { // Already sorted int[] a1 = {1,3,5,7,9; int[] b1 = {1,3,5,7,9; bubblesort(a1); assertarrayequals(a1,b1); // Reverse order int[] a2 = {9,7,5,3,1; int[] b2 = {1,3,5,7,9; bubblesort(a2); assertarrayequals(a2,b2); // Random int[] a3 = {3,1,7,9,5; int[] b3 = {1,3,5,7,9; bubblesort(a3); assertarrayequals(a3,b3); // Empty int[] a4 = {; int[] b4 = {; bubblesort(a4); assertarrayequals(a4,b4); // Singleton int[] a5 = {; int[] b5 = {; bubblesort(a5); assertarrayequals(a5,b5); 18

19 ImplementaNon: Bubble Sort A very simple sornng algorithm... public static void sort (int[] a) { for (int i = a.length; i >= 0; i--) { int j = i; while (j+1 < a.length && a[j]>a[j+1]) { int temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; j++; 19

20 A Generic Bubble Sort Unlike OCaml, Java does not support generic comparison... public static <X> void sort (X[] a) { for (int i = a.length; i >= 0; i--) { int j = i; while (j+1 < a.length && a[j] > a[j+1]) { X temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; j++; 20

21 A Generic Bubble Sort Idea: sort Comparable things: public static <X> void sort (Comparable<X>[] a) { for (int i = a.length; i >= 0; i--) { int j = i; while (j+1 < a.length && a[j].compareto(a[j+1]) > 0) { X temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; j++; interface Comparable<X> { int compareto(x o); not quite right! :- ( 21

22 A Generic Bubble Sort Fix: constrain X to be a Comparable type : public static <X extends Comparable<X>> void sort (X[] a) { for (int i = a.length; i >= 0; i--) { int j = i; while (j+1 < a.length && a[j].compareto(a[j+1]) > 0) { X temp = a[j]; a[j] = a[j+1]; a[j+1] = temp; j++; interface Comparable<X> { int compareto(x o); We ll come back to this later... 22

23 Blur 23

24 Blur 24