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Transcript of Copyright 2006 by Pearson Education 1 Building Java Programs Chapter 13: Searching and Sorting.
1Copyright 2006 by Pearson Education
Building Java Building Java ProgramsPrograms
Chapter 13: Searching and Sorting
2Copyright 2006 by Pearson Education
Chapter outline Searching and sorting in the Java class libraries
sequential and binary search sorting shuffling custom ordering with Comparators
Program efficiency algorithm analysis complexity classes and Big-Oh notation
Implementing searching and sorting algorithms binary search selection sort merge sort
3Copyright 2006 by Pearson Education
Searching and sorting Searching and sorting in the Java class in the Java class
librarieslibrariesreading: 13.1
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Sequential search Imagine searching for a particular word in an ArrayList
of words from a book:int index = words.indexOf(word);if (index >= 0) { System.out.println(word + " is word #" + index);} else { System.out.println(word + " is not found.");}
sequential search: One that examines each element of a list in sequence until it finds the target value or reaches the end of the list.
The indexOf method above uses a sequential search.
index 0 1 2 3 4 5 6 ...
value It was
the
best
of times
it ...
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Binary search algorithm binary search: An algorithm that searches for a value
in a sorted list by repeatedly eliminating half the list from consideration.
Can be written iteratively or recursively implemented in Java as method Arrays.binarySearch in java.util package
Algorithm pseudocode (searching for a value K): Start out with the search area being from indexes 0 to length-1. Examine the element in the middle of the search area.
If it is K, stop. Otherwise,
If it is smaller than K, eliminate the upper half of the search area. If it is larger than K, eliminate the lower half of the search area. Repeat the above examination.
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Using binarySearch Java provides two binary search methods:
Arrays.binarySearch (for an array)
// binary search on an array:int[] numbers = {-3, 2, 8, 12, 17, 29, 44, 58, 79};int index = Arrays.binarySearch(numbers, 29);System.out.println("29 is found at index " + index);
Collections.binarySearch (for a List)
// binary search on ArrayList with the same values:int index = Collections.binarySearch(list, 29);System.out.println("29 is found at index " + index);
Note that the values in the array / list are in sorted order. If they are not, binarySearch is not guaranteed to work
properly.
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Sorting sorting: Rearranging the values in a list into a given
order (often into their natural ascending order). One of the fundamental problems in computer science Many sorts are comparison-based (must determine order
through comparison operations on the input data)
<, >, compareTo, …
index 0 1 2 3 4 5 6 7
value 15 2 8 1 17 10 12 5
index 0 1 2 3 4 5 6 7
value 1 2 5 8 10 12 15 17
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Sorting in the class libraries Java provides two sorting methods:
Arrays.sort (for an array)// demonstrate the Arrays.sort methodString[] strings = {"c", "b", "g", "h", "d", "f", "e", "a"};System.out.println(Arrays.toString(strings));Arrays.sort(strings);System.out.println(Arrays.toString(strings));
Output:[c, b, g, h, d, f, e, a][a, b, c, d, e, f, g, h]
Collections.sort (for a List)
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Shuffling shuffling: Rearranging the elements of a list into a random order.
Java has a shuffle method for a List, Collections.shuffle:
String[] ranks = {"2", "3", "4", "5", "6", "7", "8", "9", "10", "Jack", "Queen", "King", "Ace"};String[] suits = {"Clubs", "Diamonds", "Hearts", "Spades"};ArrayList<String> deck = new ArrayList<String>();for (String rank : ranks) { // build sorted deck for (String suit : suits) { deck.add(rank + " of " + suit); }}Collections.shuffle(deck);System.out.println("Top card = " + deck.get(0));
Output (for one example run):Top card = 10 of Spades
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Custom ordering with Custom ordering with ComparatorsComparators
reading: 13.1
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Custom ordering Sometimes, the default ordering is not what you want.
Example: The following code sorts the strings in a case-sensitive order, so the uppercase letters come first.We may have wanted a case-insensitive ordering instead.
String[] strings = {"Foxtrot", "alpha", "echo", "golf", "bravo", "hotel", "Charlie", "DELTA"};Arrays.sort(strings);System.out.println(Arrays.toString(strings));
Output:[Charlie, DELTA, Foxtrot, alpha, bravo, echo, golf, hotel]
You can describe a custom sort ordering by creating a class called a comparator.
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Comparators The Comparator interface in java.util describes a
method for comparing two objects of a given type:public interface Comparator<T> { public int compare(T o1, T o2);}
Note that Comparator is a generic interface. T will be replaced by the type of object you are actually comparing.
The compare method's job is to decide the relative ordering of the two given objects and return an appropriate integer:
< 0 if o1 comes before o2, 0 if o1 and o2 are equivalent, > 0 if o1 comes after o2.
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Comparator example The following Comparator compares Strings, ignoring case:
public class CaseInsensitiveComparator implements Comparator<String> {
public int compare(String s1, String s2) { return s1.toLowerCase().compareTo( s2.toLowerCase()); }}
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Sorting with Comparators The sorting methods shown previously can also be
called with a Comparator as a second parameter. The sorting algorithm will use that comparator to order the
elements of the array or list.
String[] strings = {"Foxtrot", "alpha", "echo", "golf", "bravo", "hotel", "Charlie", "DELTA"};Arrays.sort(strings, new CaseInsensitiveComparator());System.out.println(Arrays.toString(strings));[Charlie, DELTA, Foxtrot, alpha, bravo, echo, golf, hotel]
Output:[alpha, bravo, Charlie, DELTA, echo, Foxtrot, golf, hotel]
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Efficiency efficiency: A measure of the computing resources used
by a program, such as time, memory, or disk space.
time efficiency: How quickly a program runs "Fast enough" is often relative to the task being performed.
5 minutes to render a complex 3D scene for a movie is fast. 5 minutes to search Google is slow.
Ways to measure the efficiency of a program: empirical analysis: Program the algorithm, run it, and time it. algorithm analysis: Applying techniques to mathematically
estimate the algorithm's runtime without actually coding it.
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Most individual statements take 1 unit of time to run. When multiple statements are executed sequentially,
their runtimes are added.
When statements are executed in a loop, the runtime is multiplied by the number of repetitions of the loop.
Rules for algorithm analysis
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More examples Larger statements can also occur sequentially, in which
case their runtimes are added. If larger statements are nested, their runtimes are
multiplied.
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Relative rates of growth most algorithms' runtime can be expressed as a
function of the input size N
rate of growth: measure of how quickly the graph of a function rises
goal: distinguish between fast- and slow-growing functions
we only care about very large input sizes(for small sizes, most any algorithm is fast enough)
this helps us discover which algorithms will run more quickly or slowly, for large input sizes
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Growth rate exampleConsider these graphs of functions.Perhaps each one represents an algorithm:n3 + 2n2
100n2 + 1000
• Which growsfaster?
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Range algorithm 1 Let's examine the efficiency and growth rate of three
algorithms for finding the range (difference between largest and smallest element) in an array.
First algorithm (looks at all pairs of values to find which pair is the largest and smallest): // returns the range of numbers in the given arraypublic static int range(int[] numbers) { int maxDiff = 0; for (int i = 0; i < numbers.length; i++) { for (int j = 0; j < numbers.length; j++) { int diff = Math.abs(numbers[j] - numbers[i]); maxDiff = Math.max(maxDiff, diff); } } return maxDiff;}
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Runtime of range 1
Observation: runtime seems to roughly quadruple when input size is doubled
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Range algorithm 2 The first algorithm redundantly makes each comparison
twice (when i <= j, when i > j).
Second, improved algorithm: public static int range2(int[] numbers) { int maxDiff = 0; for (int i = 0; i < numbers.length; i++) { for (int j = i + 1; j < numbers.length; j++) { int diff = Math.abs(numbers[j] - numbers[i]); maxDiff = Math.max(maxDiff, diff); } } return maxDiff;}
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Observation: about twice as fast; same growth pattern (runtime quadruples when input size is doubled)
Runtime of range 2
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Range algorithm 3 We can discover the largest and smallest values in a
single pass over the array, rather than using many nested passes over all values.
Third algorithm: public static int range3(int[] numbers) { int max = numbers[0]; int min = max; for (int i = 1; i < numbers.length; i++) { if (numbers[i] > max) { max = numbers[i]; } else if (numbers[i] < min) { min = numbers[i]; } } return max - min;}
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Runtime of range 3
Much faster than others
Runtime doubles when input size doubles
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Complexity classes Complexity classes and Big-oh notationand Big-oh notation
reading: 13.2
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complexity class: Category of algorithms' runtime based on relationship to input size.
Determined by the exponent of the most frequently executed line of code in the algorithm.
big-Oh notation: A shorthand for describing complexity classes.
Example: If the most frequently executed line of an algorithm runs approximately N3 times, we say the algorithm is "order N3" or O(N3) for short.
We are concerned with how the function grows when N is large. We are not picky about constant factors.
Big-Oh notation
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Hierarchy of Big-Oh Complexity classes in increasing order of growth:
constant time O(1) logarithmic O(log N) linear O(N) loglinear O(N log N) quadratic O(N2) cubic O(N3)
... exponential O(2N) ...
An algorithm from a lower complexity class will run much faster than one from a higher complexity class when the value of N becomes very large.
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for (int i = 0; i < N; i += c) // O(N) <statements>;
Adding to the loop counter means that the loop runtime grows linearly when compared to its maximum value N.
Loop executes its body exactly N / c times.
for (int i = 0; i < N; i *= c) // O(log N) <statements>;
Multiplying the loop counter means that the maximum value N must grow exponentially to linearly increase the loop runtime; therefore, it is logarithmic.
Loop executes its body exactly logc N times.
for (int i = 0; i < N * N; i += c) // O(N2) <statements>;
The loop maximum is N2, so the runtime is quadratic. Loop executes its body exactly (N2 / c) times.
Program loop runtimes
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Nesting loops multiplies their runtimes.for (int i = 0; i < N; i += c) { // O(N2) for (int j = 0; j < N; i += c) { <statements>; }}
Loops in sequence add together their runtimes, which means the loop set with the larger runtime dominates.for (int i = 0; i < N; i += c) { // O(N) <statements>;}for (int i = 0; i < N; i += c) { // O(N log N) for (int j = 0; j < N; i *= c) { <statements>; }}
More loop runtimes
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Loop runtime problems Approximate the value of the variable sum after the
following code fragment, as an expression in terms of input size n. Use Big-Oh notation to describe the algorithm's overall runtime.int sum = 0;for (int i = 1; i <= N; i *= 2) { sum++;} for (int i = 1; i <= 1000; i++) { sum++; sum++;}
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Loop runtime problems Approximate the value of the variable sum after the
following code fragment, as an expression in terms of input size n. Use Big-Oh notation to describe the algorithm's overall runtime.int sum = 0;for (int i = 1; i <= N; i++) { for (int j = 1; j <= i / 2; j += 2) { sum++; }}
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Implementing binary Implementing binary searchsearch
reading: 13.3
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Recall: binary search binary search: An algorithm that searches for a value
in a sorted list by repeatedly eliminating half the list from consideration.
Can be written iteratively or recursively
Algorithm pseudocode (searching for a value K): Start out with the search area being from indexes 0 to length-1. Examine the element in the middle of the search area.
If it is K, stop. Otherwise,
If it is smaller than K, eliminate the upper half of the search area. If it is larger than K, eliminate the lower half of the search area. Repeat the above examination.
38Copyright 2006 by Pearson Education
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min
mid (too big!)
max
searching for value 16
Binary search example
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Binary search example
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searching for value 16
Binary search example
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Binary search pseudocodebinary search array a for value i: if all elements have been searched, result is -1. examine middle element a[mid]. if a[mid] equals i, result is mid. if a[mid] is greater than i, binary search lower half of a for i. if a[mid] is less than i, binary search upper half of a for i.
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Binary search code The following code implements binary search:
// Returns an index at which the target// appears in the given input array, or -1 if not found.// pre: array is sorted.public static int binarySearch(int[] numbers, int target) { int min = 0; int max = numbers.length - 1;
while (min <= max) { int mid = (max + min) / 2; if (numbers[mid] == target) { return mid; // found it! } else if (numbers[mid] < target) { min = mid + 1; // too small } else { // numbers[mid] > target max = mid - 1; // too large } }
return -1; // not found}
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Runtime of binary search How do we analyze the runtime of binary search?
The algorithm's runtime is dominated by a while loop. Each pass of that loop cuts the search space in half.
How many times does this division in half take place? 2repetitions N repetitions log2N
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Implementing Implementing selection sortselection sort
reading: 13.3
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Selection sort selection sort: orders a list of values by repetitively
putting a particular value into its final position
more specifically: find the smallest value in the list switch it with the value in the first position find the next smallest value in the list switch it with the value in the second position repeat until all values are in their proper places
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Index0 1 2 3 4 5 6 7
Value27 63 1 72 64 58 14 9
1st pass1 63 27 72 64 58 14 9
2nd pass1 9 27 72 64 58 14 63
3rd pass1 9 14 72 64 58 27 63
…
Selection sort example 2
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Selection sort code The following code implements selection sort:
// places elements of the given array into sorted orderpublic static void selectionSort(int[] a) { for (int i = 0; i < a.length - 1; i++) { // find index of smallest element int smallest = i; for (int j = i + 1; j < a.length; j++) { if (a[j] < a[smallest]) { smallest = j; } }
swap(a, i, smallest); // swap smallest to front }}
public static void swap(int[] list, int i, int j) { int temp = list[i]; list[i] = list[j]; list[j] = temp;}
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Observation: runtime roughly quadruples when input size is doubled
makes sense because the code performs two nested loops over the array of size N
its runtime should be O(N2)
Selection sort runtime
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Sorting practice problem Consider the following array of int values:
[22, 11, 34, -5, 3, 40, 9, 16, 6]
Perform the selection sort algorithm on this data.Write the contents of the array after each pass of the outermost loop of selection sort until the array is sorted.
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Merge sort merge sort: orders a list of values by recursively
dividing the list in half until each sub-list has one element, then recombining
Invented by John von Neumann in 1945 relies on observation that if you have two sorted lists of values,
they can be combined into a single larger sorted list quickly
merge sort pseudocode: split the array into two halves. merge sort the left half (recursively). merge sort the right half (recursively). combine the two halves into a sorted whole.
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674523 14 6 3398 42
4523 1498 676 33 42
45 142398
2398 45 14
676 33 42
676 33 42
23 98 4514 676 4233
14 23 45 98 6 33 42 67
6 14 23 33 42 45 67 98
Merge sort example
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13 6 21 18 9 4 8 20
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Merge sort example 2
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merge operation: Given two sorted arrays, merge operation produces a sorted
array with all the elements of the two arrays
A 6 13 18 21 B 4 8 9 20
C 4 6 8 9 13 18 20 21
Running time of merge : O(N), where N is the number of elements in the merged array.
When merging two sorted parts of the same array, we'll need a temporary array to store the merged whole.
Merging two sorted arrays
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Split into halves code The following code splits an array into its two halves:
// Returns the first half of the given array.public static int[] leftHalf(int[] array) { int size1 = array.length / 2; int[] left = new int[size1]; for (int i = 0; i < size1; i++) { left[i] = array[i]; } return left;}
// Returns the second half of the given array.public static int[] rightHalf(int[] array) { int size1 = array.length / 2; int size2 = array.length - size1; int[] right = new int[size2]; for (int i = 0; i < size2; i++) { right[i] = array[i + size1]; } return right;}
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Merge code The following code implements the merge operation
// pre : result is empty; left/right are sorted// post: result contains merged sorted lists.public static void merge(int[] result, int[] left, int[] right) { int i1 = 0; // index into left array int i2 = 0; // index into right array for (int i = 0; i < result.length; i++) { if (i2 >= right.length || (i1 < left.length && left[i1] <= right[i2])) { result[i] = left[i1]; // take from left i1++; } else { result[i] = right[i2]; // take from right i2++; } }}
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Merge sort code The following code implements the overall merge sort:
// Places elements of given array into sorted order// using the merge sort algorithm.public static void mergeSort(int[] array) { if (array.length > 1) { // split array into two halves int[] left = leftHalf(array); int[] right = rightHalf(array);
// recursively sort the two halves mergeSort(left); mergeSort(right);
// merge the sorted halves into a sorted whole merge(array, left, right); }}