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Java Programming: From Problem Analysis to Program Design, 3e Chapter 10 Applications of Arrays.

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Presentation on theme: "Java Programming: From Problem Analysis to Program Design, 3e Chapter 10 Applications of Arrays."— Presentation transcript:

1 Java Programming: From Problem Analysis to Program Design, 3e Chapter 10 Applications of Arrays

2 class Arrays (API)API Contains a number of static methods that can be used to work with arrays. The various methods accept arguments to work with arays: –Arrays.sort(double[] arr) : sorts the array into increasing order –Arrays.equals(double[] a, double[] b) : returns true if the arrays are identical in size and values –Arrays.fill(int[] arr, int n) : fills the array with ‘n’s –Arrays.binarySearch(int[], -23) : returns the index of the first time that -23 is found in the array. Returns -1 if the number is not found. Note that most of these methods are overloaded. For example, there is a Arrays.fill() method to fill String, doubles, chars, etc. These methods are also overloaded to accept arrays of Objects. This allows you to do things like compare two StudentRecord objects: Arrays.equals(sr1, sr2); –Of course, this does require that the equals method has been overloaded in your StudentRecord clas. Now let’s turn to the class ArrayList

3 *** Collections: ArrayList Recall how the size of an array is declared when you first instantiate the array. This is an important and significant limitation. If your array is full and you then realize you need to add an element to your array, you would be required to do something like create a new array and then copy all elements of the old array to the new array. The moment you need to add yet another element, you’d have to repeat this process. This is very wasteful in terms of computing resources. In the bad-old days, programmers would try to get around this by declaring an array of size “much_bigger_than_I_will_ever_need” and simply keep track of the number of elements currently being used by the array. Languages that support “collections” such as Java, give us a much nicer tool. In Java, we can use a class called ArrayList (API). This class allows us to store arrays of objects, add elements, remove elements, and search for elements (and do a couple of other things as well).API This class does not use array syntax with square brackets. Instead, it uses methods. NOTE: ArrayList can NOT be used with primitive data types (int, double, etc). Instead, you can use the wrapper classes Integer, Double, etc.

4 Some examples: ArrayList names = new ArrayList (); ArrayList students = new ArrayList (); ArrayList numbers = new ArrayList (); StudentRecord s1 = new StudentRecord(); students.add(s1); names.add("Lisa"); names.add("Max"); if ( names.contains("Max") ) {... //returns true names.remove("Alan"); //nothing will happen

5 public static void main(String[] args) { ArrayList words = new ArrayList (); words.add("hello"); words.add("goodbye"); words.add("test again"); for (int i=0; i<words.size(); i++) System.out.print( words.get(i) + "\n"); } //end main Simple ArrayList Example Note the method size() instead of the field ‘length’ that we typically use with arrays. Note the method get() instead of the usual arrayName[] we used with arrays.

6 *** Study and play with the file ArrayListPrac.java on the class page. Note that an ArrayList can –usually does– hold objects. The example here holds Strings, but you can easily have lists of StudentRecords, Die, etc, etc Practice by creating a list of, say, Die objects and then outputting them, searching through them, etc API is here. Note important methods such as: –get(int)  returns an object at location of the int. E.g. get(0) will return the first item stored in the list e.g. System.out.println( list.get(2) ); will output the object at the 3 rd position in the list. –size()  returns the size of the list

7 Note: Your book spends some time discussing a technique using something called Vectors. This is a relatively dated collections resource. You can skip this section and focus on the ArrayList instead.

8 Algorithms We’ve spent a good amount of time over the past 1.5 courses learning about the syntax of programming. However, most of the challenge and, yes, the fun of programming lies in coming up with creative and efficient ways of solving problems. Today we will spend some time trying to solve the problem of putting a list into some kind of order. We will look at a couple of different approaches to solving the problem (known as an algorithm) and will attempt to gradually improve our algorithms in order to solve our problem as efficiently as possible.

9 List Processing List: a set of values of the same type –e.g. an array Basic operations performed on a list –Search list for given item –Insert item in list –Delete item from list –Sort list

10 Writing a method to search a list: Necessary information for our method: –Array containing the list –Length of the list Remember: Not every array is completely filled –Item for which you are searching After search completed –If item found, return an int representing the location in array –If item not found, return -1

11 Why not return a boolean? In our discussion of the method (previous slide) note that the method returns an int. Wouldn’t it make more sense to have a method that looks for an item in an array return a boolean? Why an int?? The reason is that frequently when searching for an item in an array, it’s not enough to simply know if the item is present, we also want to know where in the array the item can be found. For this reason, it was decided to write the method so that not only does the method tell us if the item is indeed present, the method also tells us where the item is located. This is why the method does not simply return a boolean, but rather, an int. This example also serves to illustrate how putting some thought into the design of your methods can make them much more useful.

12 Suppose that you want to determine whether 27 is in the list First compare 27 with list[0] ; that is, compare 27 with 35 Because list[0] ≠ 27, you then compare 27 with list[1] Because list[1] ≠ 27, you compare 27 with the next element in the list Because list[2] = 27, the search stops This search is successful Searching for an item

13 Search (continued) Let’s now search for 10 (not present in the array) The search starts at the first element in the list; that is, at list[0] Proceeding as before, we see that this time the search item, which is 10, is compared with every item in the list Eventually, no more data is left in the list to compare with the search item; this is an unsuccessful search

14 Search: the code

15 Search - continued ( Many ways to skin a cat)

16 Using a while (or a for ) loop, the definition of the method seqSearch can also be written without the break statement as: Search (continued)

17 * An Aside: break and continue The ‘break’ statement when executed in a loop (while, for, do), causes flow to immediately exit the loop. Execution continues immediately after the body of the loop. continue: Also is used within loops. However, instead of exiting the loop, flow skips the remainder of the loop’s body. It then returns to the top of the loop for another iteration. ** Some programmers feel that break and continue are poor examples of properly structured programming. We won’t delve into these issues here, but do be aware that use of break / continue is somewhat contraversial.

18 Sorting Our Lists (arrays) Consider the simple search we’ve just talked about. For an array of 1000 items, a search will require a minimum of 1 comparison (if the item is found in the very first element of the array) to a maximum of 1000 comparisons (the item is not present in the array). Statistically, for an array of 1000 items, searches will require an average of 500 comaprisons. When you consider that some arrays can be much, much larger than 1000 items and also that we may sometimes need to do many, many searches, this can add up to a very slow application. If we know that we are going to need to do lots of searching, it would be a highly useful investment of time to sort our array in order (e.g. alphabetic, or numeric). Sorting arrays is an important technique in programming, so we will spend some time on it now.

19 Sorting a List There are many sorting algorithms out there, we will begin with one known as the “bubble sort”. Bubble sort –Go ahead and read the points here, but it will make much more sense after studying the illustrations on the following slides –Suppose list[0...n - 1] is a list of n elements, indexed 0 to n - 1 –We want to rearrange; that is, sort, the elements of list in increasing order –The bubble sort algorithm works as follows: In a series of n - 1 iterations, the successive elements, list[index] and list[index + 1] of list are compared If list[index] is greater than list[index + 1], then the elements list[index] and list[index + 1] are swapped, that is, interchanged

20 Bubble Sort

21 Bubble Sort (continued)

22

23 It is known that for a list of length n, on average bubble sort makes n(n – 1) / 2 key comparisons and about n(n – 1) / 4 item assignments Therefore, if n = 1000, then to sort the list bubble sort makes about 500,000 key comparisons and about 250,000 item assignments. This amounts to ¾ of a million operations! In other words, this is hardly an efficient way of sorting! Inefficiency = slow-running application Bubble Sort (continued)

24 Selection Sort Algorithm The “selection sort” is another sorting algorithm that has the advantage of being much more efficient than the bubble sort algorithm. List is sorted by selecting list element and moving it to its proper position Algorithm finds position of smallest element and moves it to top of unsorted portion of list Repeats process above until entire list is sorted

25 Selection Sort Algorithm - Pseudocode for (index = 0; index < listLength-1; index++) { a. find the location, smallestIndex, of the smallest element in the list array. b. Swap the smallest element with list[index]. That is, swap list[smallestIndex] with list[index]. } Take a look at the code example on this slide. Note that is a combination of programming syntax and English. This is known as “pseudocode”. Pseudocode technique is one of the VERY BEST ways you can begin trying to come up with an algorithm to solve a problem. You should be familiar with this term, and if I were to ask you to give me an example of pseudocode in homework or exams, you should be able to do so. The pseudocode used by the textbook for the selection sort is replicated here:

26 Selection Sort (continued)

27

28 public static void selectionSort(int[] list, int listLength) { int index; int smallestIndex; int minIndex; int temp; for (index = 0; index < listLength – 1; index++) { //find the location of the smallest element in the array smallestIndex = index; for (minIndex = index + 1; minIndex < listLength; minIndex++) if (list[minIndex] < list[smallestIndex]) smallestIndex = minIndex; //swap the smallest element in the array with list[index] temp = list[smallestIndex]; list[smallestIndex] = list[index]; list[index] = temp; } Selection Sort (continued)

29 It is known that for a list of length n, on an average selection sort makes n(n – 1) / 2 key comparisons and 3(n – 1) item assignments Therefore, if n = 1000, then to sort the list selection sort makes about 500,000 key comparisons and about 3000 item assignments In other words, compare an average of 500,000 assignments in the selection sort with 750,000 in the bubble sort. However, this is still quite high… Selection Sort (continued)

30 Insertion Sort for (firstOutOfOrder=1; firstOutOfOrder<listLength; firstOutOfOrder++) { if ( list[firstOutOfOrder] is less than list[firstOutOfOrder-1] { 1. copy list[firstOutOrOrder] into temp 2. initialize location to firstOutOrOrder 3. do { a. move list[location-1] one array slot down b. decrement location by 1 to consider the next element sorted of the portion of the array } while ( location>0 && the element in the upper list at location-1 is greater than temp 4. copy temp into list[location] This is another algorithm that attempts to improve, yet again, the efficiency of our sort. Here is the pseudocode as given in your textbook:

31 Insertion Sort The insertion sort algorithm sorts the list by moving each element to its proper place

32 Insertion Sort (continued)

33

34

35 public static void insertionSort(int[] list, int listLength) { int firstOutOfOrder, location; int temp; for (firstOutOfOrder = 1; firstOutOfOrder < listLength; firstOutOfOrder++) if (list[firstOutOfOrder] < list[firstOutOfOrder - 1]) { temp = list[firstOutOfOrder]; location = firstOutOfOrder; do { list[location] = list[location - 1]; location--; } while(location > 0 && list[location - 1] > temp); list[location] = temp; } } //end insertionSort Insertion Sort (continued)

36 It is known that for a list of length n, on average, the insertion sort makes (n 2 + 3n – 4) / 4 key comparisons and about n(n – 1) / 4 item assignments Therefore, if n = 1000, then to sort the list, the insertion sort makes about 250,000 key comparisons and about 250,000 item assignments Insertion Sort (continued)

37 Searching Now that we have sorted our lists, let’s return to a discussion of searching. Once we have an ordered list of items, we can significantly improve the performance (efficiency) of our searching algorithms. The two best known algorithms are the sequential search and the binary search.

38 Sequential Search This search begins by looking at the first item in the array, then the second, then the third, etc, etc. The advantage to having a sorted array is that the moment we get to an item that is greater than the item we are searching for, we know that the item must not be in the list and we can stop iterating through the array. Eg: We have an array: { 3 9 12 17 23 } and are looking for the number 8. The moment we hit the second element in our array (the 9), we know that 8 must not be present and can stop searching. The code for a sequential search is shown on the next slide.

39 Sequential Ordered Search public static int seqOrderedSearch(int[] list, int listLength, int searchItem) { int loc; //Line 1 boolean found = false; //Line 2 for (loc = 0; loc < listLength; loc++) //Line 3 if (list[loc] >= searchItem) //Line 4 { found = true; //Line 5 break; //Line 6 } if (found) //Line 7 if (list[loc] == searchItem) //Line 8 return loc; //Line 9 else //Line 10 return -1; //Line 11 else //Line 12 return -1; //Line 13 }

40 Binary Search Can only be performed on a sorted list We start half-way through our sorted list. If the the item in the half-way array index is lower than the item we are looking for, we continue searching, but in the upper (later) half of the array only. Similarly, if the item in half-way array index is greater than our search item, we can continue looking in the lower half of the array only. We then repeat the process, dividing the array in half each time. Sometimes called the “divide and conquer technique” Illustrations follow…

41 Binary Search Algorithm Search item is compared with middle element of list If search item < middle element of list, search is restricted to first half of the list If search item > middle element of list, search second half of the list If search item = middle element, search is complete

42 Binary Search Algorithm (continued) Determine whether 75 is in the list

43 Binary Search Algorithm (continued)

44 public static int binarySearch(int[] list, int listLength, int searchItem) { int first = 0; int last = listLength - 1; int mid; boolean found = false; while (first <= last && !found) { mid = (first + last) / 2; if (list[mid] == searchItem) found = true; else if (list[mid] > searchItem) last = mid - 1; else first = mid + 1; } if (found) return mid; else return –1; } //end binarySearch Binary Search Algorithm (continued)

45

46

47 Performance of the Binary Search

48 Performance of the Binary Search (continued)

49 Suppose that L is a list of size 1000000 Since 1000000  1048576 = 220, it follows that the while loop in binary search will have at most 21 iterations to determine whether an element is in L Every iteration of the while loop makes two key (that is, item) comparisons Performance of the Binary Search (continued)

50 To determine whether an element is in L, binary search makes at most 42 item comparisons –On the other hand, on average, a sequential search will make 500,000 key (item) comparisons to determine whether an element is in L In general, if L is a sorted list of size n, to determine whether an element is in L, the binary search makes at most 2log2n + 2 key (item) comparisons

51 Performance of the Binary Search (in English) Ultimately, for an array of 1000 items, the maximum number of operations turns out to be 11 ! Compare this with a maximum of 1000 in a sequential search.

52 String Class – Second look: Now that you have a pretty good if basic understanding of objects and arrays, you will hopefully have little problems understanding most or all of the methods in the String class. The slides that follow are a partial API of the String class. While we will not cover these in any detail in class, you are expected to be able to apply them if called upon (homework, exams, etc)

53 Additional String Methods

54 Additional String Methods (continued)

55 Additional String Methods

56

57 Effects of Some String Methods

58 Chapter Summary Lists Searching lists –Sequential searching –Sequential searching on an order list –Binary Search Sorting lists –Bubble Sort –Selection Sort –Insertion Sort

59 Chapter Summary (continued) Programming examples The class Vector –Members of the class Vector The class String –Additional methods of the class String


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