Searching in Arrays in Java

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Searching in Arrays
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The Arrays class provides enough overloaded versions of the binarySearch() method to search in practically any type of array that is sorted The discussion on searching in lists (p 840) is also applicable to searching in arrays
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CHAPTER 15: COLLECTIONS AND MAPS
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The methods below return the index to the key in the sorted array, if the key exists If not, a negative index is returned, corresponding to (-insertion point)-1, where insertion point is the index of the element where the key would have been found if it had been in the array In case there are duplicate elements equal to the key, there is no guarantee which duplicate s index will be returned The elements and the key must be mutually comparable The bounds, if specified in the methods below, define a half-open interval The search is then confined to this interval
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int binarySearch(type[] array, type key) int binarySearch(type[] array, int fromIndex, int toIndex, type key)
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Permitted type for elements include byte, char, double, float, int, long, short, and Object In the case where an array of objects is passed as argument, the objects must be sorted in natural ordering, as defined by the Comparable interface
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<E> int binarySearch(E[] array, E key, Comparator< super E> c) <E> int binarySearch(E[] array, int fromIndex, int toIndex, E key, Comparator< super E> c)
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The two generic methods above require that the array is sorted according to the total ordering dictated by the comparator In particular, its elements are mutually comparable according to this comparator The comparator must be equivalent to the one that was used for sorting the array, otherwise the results are unpredictable The experiment from p 840 with a list of strings is now repeated with an array of strings, giving identical results In the code below the return value -3 indicates that the key would have been found at index 2 had it been in the list
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// Sorted String array (natural order): [Bigfoot, big, Arrayssort(strArray); // Search in natural order: outprintln(ArraysbinarySearch(strArray, "bigger")); outprintln(ArraysbinarySearch(strArray, "bigfeet")); outprintln(ArraysbinarySearch(strArray, "bigmouth")); bigger, biggest]
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// Successful: 2 // Unsuccessful: -3 // Unsuccessful: -5
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Results are unpredictable if the array is not sorted or the ordering used in the search is not the same as the sort ordering Searching in the strArray using reverse natural ordering when the array is sorted in natural ordering gives the wrong result:
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outprintln(ArraysbinarySearch(strArray, "bigger", CollectionsreverseOrder())); // -1 (INCORRECT)
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A ClassCastException is thrown if the key and the elements are not mutually comparable:
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outprintln(ArraysbinarySearch(strArray, 4)); // Key: 4 => ClassCastException
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However, this incompatibility is caught at compile time in the case of arrays with primitive values:
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// Sorted int array (natural order): [1, 3, 5, 7] outprintln(ArraysbinarySearch(intArray, 45));// Key: 45 => Compile time error!
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1511: WORKING WITH COLLECTIONS
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Creating List Views of Arrays
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The asList() method in the Arrays class and the toArray() method in the Collection interface provide the bidirectional bridge between arrays and collections The asList() method of the Arrays class creates List views of arrays
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<E> List<E> asList(E elements)
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Returns a fixed-size list view backed by the array corresponding to the vararg argument elements Changes to the List view reflect in the array, and vice versa The List is said to be backed by the array The List size is equal to the array length and cannot be changed The code below illustrates using the asList() method The list1 is backed by the array1 at (1) The list2 is backed by an implicit array of Integers at (2) An array of primitive type cannot be passed as argument to this method, as evident by the compile time error at (3)
Integer[] array1 = new Integer[] {9, 1, 1}; int[] array2 = new int[] {9, 1, 1}; List<Integer> list1 = ArraysasList(array1); List<Integer> list2 = ArraysasList(9, 1, 1); // List<Integer> intList3 = ArraysasList(array2);
// (1) An array of Integers // (2) A var-arg list // (3) Compile-time error!
Various operations on the list1 show how changes are reflected in the backing array1 Elements cannot be added to the list view (shown at (5)), and elements cannot be removed from a list view (shown at (10)) An UnsupportedOperationException is thrown in both cases An element at a given position can be changed, as shown at (6) The change is reflected in the list1 and the array1, as shown at (7a) and (7b), respectively A sublist view is created from the list1 at (8), and sorted at (11) The changes in the sublist1 are reflected in the list1 and the backed array1
Systemoutprintln(list1); // (4) [9, 1, 1] // list1add(10); // (5) UnsupportedOperationException list1set(0, 10); // (6) Systemoutprintln(list1); // (7a) [10, 1, 1] Systemoutprintln(ArraystoString(array1)); // (7b) [10, 1, 1] List<Integer> sublist1 = list1subList(0, 2); // (8) Systemoutprintln(sublist1); // (9) [10, 1] // sublist1clear(); // (10) UnsupportedOperationException Collectionssort(sublist1); // (11) Systemoutprintln(sublist1); // (12a) [1, 10] Systemoutprintln(list1); // (12b) [1, 10, 1] Systemoutprintln(ArraystoString(array1)); // (12c) [1, 10, 1]
The code below shows how duplicates can be eliminated from an array using these two methods:
String[] jiveArray = new String[] {"java", "jive", "java", "jive"}; Set<String> jiveSet = new HashSet<String>(ArraysasList(jivearray)); String[] uniqueJiveArray = jiveSettoArray(new String[0]); // (1) // (2)