What is array index in java

Java java array what is index 1

For example: this SO Q&A Solution 2: First, having two parallel arrays with elements in corresponding indexes indicates you might want to change the data structure to a single array with a container object containing two fields. In this case startAddress is the value in charArray, index is the number 1 and size per element is the size of an individual char (1 Byte) so it gives us the memory address of 8.

Accessing the nth index of array in o(1)

Indexing an array in O(1) is possible because the calculation to access the data you want is done in one step.

How this is done is quite simple, say you create an array of 10 integers.

What is happening here is that you are asking for a block of memory that is big enough to contain 10 integers (In Java they are 32 bit, so 10 * 4 Bytes long).

When the memory has been reserved, the memory address at the beginning of this block is given back and put into the intArray variable (a variable containing an address/reference to something else is called a pointer).

To make this easier to understand you can picture memory (RAM) as a long list of single byte blocks that each have an address.

Data: [ 00000000 ] [ 11111111 ] [ 00001111 ] Address: 7 8 9 

When you use an index to get access to a part of the array, what is happening is that the index is used to calcuate the memory address of the block you want on the fly.

e.g. using the example above

char[] charArray = new char[3]; // This returns memory address 7 charArray[1] = 'b'; // Here the memory address is calcuated with the formula // startAddress + index * Size Per Element 

In this case startAddress is the value in charArray, index is the number 1 and size per element is the size of an individual char (1 Byte) so it gives us the memory address of 8.

As said by @Nullman in the comments, and at least when refering to Python, list s are accessed in O(1).

Java: What is index based access of arrays and why are, ArrayList is backed by an array. Typically, addressing something by index is assumed to be O (1), i.e., in constant time. In assembly, you can index into a memory location in constant time, as there are certain LOAD instructions that take an optional index in addition to a base address. Arrays are usually allocated … Usage exampleint a = myArray[9];Feedback

Sorting an array and keep the index [duplicate]

Sort both data structures in parallel.

Or make the reference index (to reference the other data structure) an internal field of each element, so that each element maintains the same «reference index» even if the order of the elements is changed by sorting.

Or better yet, just give each element an internal field that is just a direct reference variable to the element in the other data structure. (unless you need the raw index itself for some reason and not just a reference to the associated object / value)

First, having two parallel arrays with elements in corresponding indexes indicates you might want to change the data structure to a single array with a container object containing two fields. This container object would then implement Comparable interface.

But, sticking with what you say, one approach could be:

/** * Sorts parallel arrays in-place. Sorted by the first array and updating * all other arrays to match. * Uses the natural sorting of the objects. * All arrays must be the same length. * * @param keys the values used to sort, may be duplicate * * @param otherArrays the arrays to have reordered to match the sorting of * the keys array. * * @exception IllegalArgumentException if any of otherArrays have a length * different that the keys array. */ public static > void sortParallelArrays( E[] keys, Object[] . otherArrays ) < int numKeys = keys.length; int numOtherArrays = otherArrays.length; for(Object[] otherArray : otherArrays) < if(otherArray.length != numKeys) < throw new IllegalArgumentException("Mismatched array lengths"); >> // A list of all indexes per key // This also does the sorting within the TreeMap using natural ordering SortedMap> originalIndexesByKey = new TreeMap>(); // Populate the map for(int i = 0; i < numKeys; i++) < E key = keys[i]; ListoriginalIndexes = originalIndexesByKey.get(key); if(originalIndexes == null) < // Optimization for the non-duplicate keys originalIndexesByKey.put(key, Collections.singletonList(i)); >else < if(originalIndexes.size() == 1) < // Upgrade to ArrayList now that know have duplicate keys originalIndexes = new ArrayList(originalIndexes); originalIndexesByKey.put(key, originalIndexes); > originalIndexes.add(i); > > // Store back to keys and sort other arrays in a single traversal Object[][] sortedOtherArrays = new Object[numOtherArrays][numKeys]; int pos = 0; for(Map.Entry> entry : originalIndexesByKey.entrySet()) < E key = entry.getKey(); for(int index : entry.getValue()) < keys[pos] = key; for(int ooIndex = 0; ooIndex < numOtherArrays; ooIndex++) < sortedOtherArrays[ooIndex][pos] = otherArrays[ooIndex][index]; >pos++; > > assert pos == numKeys : "Arrays should be full"; // Copy back to original arrays for in-place sort for(int ooIndex = 0; ooIndex < numOtherArrays; ooIndex++) < System.arraycopy( sortedOtherArrays[ooIndex], 0, otherArrays[ooIndex], 0, numKeys); >> 

This is not the most memory efficient strategy, but isn’t much code.

The time complexity isn’t too bad. Looks something like O((M+1)*N*log(N)) , where M is the number of otherArrays and N is the number of keys. No crazy worst-case issues, at least.

Java — sorting an array and keep the index, Sorted by: 1. Sort both data structures in parallel. Or make the reference index (to reference the other data structure) an internal field of each element, so that each element maintains the same «reference index» even if the order of the elements is changed by sorting. Or better yet, just give each element …

Java — Initial value of an array of objects

x is an array of 20 Job objects, all of which are initialized to null . If you want to initialize each one to be a new object you can use a for loop:

Each index in an array references null until it is initialized to another value.

will hold 2 Job references, but they will both be null . That is until you initialize them

jobs[0] = new Job(); jobs[1] = new Job(); 

Note that in this case, you’ve declared x as an instance field of Job , so each new Job object you create will have a Job array with 20 null Job references.

Java — How can I avoid ArrayIndexOutOfBoundsException, java.lang.IndexOutOfBoundsException: Index: 1, Size: 1. Where Index is the index that you requested that does not exist and Size is the length of the structure you were indexing into. As you can see a Size: 1 means the only valid index is 0 and you were asking for what was at index 1. For example, if you have …

Источник

Java Arrays

Arrays are used to store multiple values in a single variable, instead of declaring separate variables for each value.

To declare an array, define the variable type with square brackets:

We have now declared a variable that holds an array of strings. To insert values to it, you can place the values in a comma-separated list, inside curly braces:

To create an array of integers, you could write:

Access the Elements of an Array

You can access an array element by referring to the index number.

This statement accesses the value of the first element in cars:

Example

String[] cars = ; System.out.println(cars[0]); // Outputs Volvo 

Note: Array indexes start with 0: [0] is the first element. [1] is the second element, etc.

Change an Array Element

To change the value of a specific element, refer to the index number:

Example

Example

String[] cars = ; cars[0] = "Opel"; System.out.println(cars[0]); // Now outputs Opel instead of Volvo 

Array Length

To find out how many elements an array has, use the length property:

Example

String[] cars = ; System.out.println(cars.length); // Outputs 4 

COLOR PICKER

Report Error

If you want to report an error, or if you want to make a suggestion, do not hesitate to send us an e-mail:

Thank You For Helping Us!

Your message has been sent to W3Schools.

Top Tutorials

Top References

Top Examples

Get Certified

W3Schools is optimized for learning and training. Examples might be simplified to improve reading and learning. Tutorials, references, and examples are constantly reviewed to avoid errors, but we cannot warrant full correctness of all content. While using W3Schools, you agree to have read and accepted our terms of use, cookie and privacy policy.

Источник