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2.1 An array component can be any type: primitive, reference, or array. An array element is a component that is not itself an array type. So in a two-dimensional array a[][], the components of a[] are its row arrays, and the elements of a[][] are double variables. A multidimensional array is one that has more than one index. A Java array has only one index variable. However, since a components indexed by that variable can itself be an array (with an index), the original array appears to have more than one index. An ArrayIndexOutOfBoundsException object is an exception that gets thrown whenever a value less than 0 or greater than or equal to the array s length is attempted to be used as an index on the array. This give the programmer some control over the consequences of such a run-time error. In languages such as C++, such a run-time error normally causes the program to crash. An array index can have type byte, char, short, or int. The Array class cannot be instantiated because its constructor is declared private. The simplest way to print an array of objects is to pass it to the Arrays.toList() method which produces a List object that can be printed directly with the System.out.println() method. The binary search will probably not work unless the sequence is sorted first. If the sequential search is applied to an element that occurs more than once in an array, it will return the index of the one that is closest to the beginning of the array. If the binary search is applied to an element that occurs more than once in an array, it could return the index of any one of them. It depends upon how close their indexes are to multiples of midpoints of subintervals. For example, if the binary search is applied in an array of 10,000 elements, searching for a value that is repeated at locations 0 99, the search would return the index 77 on the 7th iteration.
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public class TestFill { public static void main(String[] args) { Object[] a = new Object[4]; Arrays.fill(a, new Date()); ch02.ex02.DuplicatingArrays.print(a); Arrays.fill(a, 22); ch02.ex02.DuplicatingArrays.print(a); Arrays.fill(a, "Yo!"); ch02.ex02.DuplicatingArrays.print(a); }
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The sequential search runs in linear time, which means that the time is proportional to the number of elements. So an array with twice as many elements would take twice as long to process: 20 ms. The binary search runs in logarithmic time, so squaring the size of the problem should only double its running time. So an array with 10002 elements would take twice as long to process: 10 ms. The interpolation search runs in hyperlogarithmic time, so squaring the size of the problem should have no appreciable effect on its running time. So an array with 1,000,000 elements would also take about 2 ms to process.
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public class TestBinarySearch { private static final int SIZE = 10000; private static final int START = 0; private static final int RANGE = 10000; private static Random random = new Random(); private static int count = 0; public static void main(String[] args) { int[] a = new int[SIZE]; load(a, START, RANGE); Arrays.sort(a); search(a, random.nextInt(10000)); System.out.println(count + " iterations"); } public static void load(int[] a, int start, int range) { for (int i = 0; i < a.length; i++) { a[i] = start + random.nextInt(range); // random 5-digit numbers } } public static int search(int[] a, int x) { int lo = 0; int hi = a.length; while (lo < hi) { ++count; int i = (lo + hi)/2; if (a[i] == x) { return i; } else if (a[i] < x) { lo = i+1; } else { hi = i; } } return -1; } }
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boolean isSorted(int[] a) { if (a.length < 2) { return true; } for (int i = 1; i < a.length; i++) { if (a[i] < a[i-1]) { return false; } } return true; }
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