Monthly Archives: January 2014

manually install JAVA SDK

/etc/profile
~/.bashrc

JAVA_HOME=/
PATH=$PATH:$HOME/bin
export JAVA_HOME
export PATH

Can’t OPEN Eclipse in Ubuntu 12.04 – java.lang.UnsatisfiedLinkError: Could not load SWT library

http://stackoverflow.com/questions/10970754/cant-open-eclipse-in-ubuntu-12-04-java-lang-unsatisfiedlinkerror-could-not-l

try with the following command

  • 64 Bits System: ln -s /usr/lib/jni/libswt-* ~/.swt/lib/linux/x86_64/
  • 32 Bits System: ln -s /usr/lib/jni/libswt-* ~/.swt/lib/linux/x86/
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Java Object Sorting Example (Comparable And Comparator)

http://www.mkyong.com/java/java-object-sorting-example-comparable-and-comparator/

In this tutorial, it shows the use of java.lang.Comparable and java.util.Comparator to sort a Java object based on its property value.

1. Sort an Array

To sort an Array, use the Arrays.sort().

	String[] fruits = new String[] {"Pineapple","Apple", "Orange", "Banana"}; 

	Arrays.sort(fruits);

	int i=0;
	for(String temp: fruits){
		System.out.println("fruits " + ++i + " : " + temp);
	}

Output

fruits 1 : Apple
fruits 2 : Banana
fruits 3 : Orange
fruits 4 : Pineapple

2. Sort an ArrayList

To sort an ArrayList, use the Collections.sort().

	List<String> fruits = new ArrayList<String>();

	fruits.add("Pineapple");
	fruits.add("Apple");
	fruits.add("Orange");
	fruits.add("Banana");

	Collections.sort(fruits);

	int i=0;
	for(String temp: fruits){
		System.out.println("fruits " + ++i + " : " + temp);
	}

Output

fruits 1 : Apple
fruits 2 : Banana
fruits 3 : Orange
fruits 4 : Pineapple

3. Sort an Object with Comparable

How about a Java Object? Let create a Fruit class:

public class Fruit{

	private String fruitName;
	private String fruitDesc;
	private int quantity;

	public Fruit(String fruitName, String fruitDesc, int quantity) {
		super();
		this.fruitName = fruitName;
		this.fruitDesc = fruitDesc;
		this.quantity = quantity;
	}

	public String getFruitName() {
		return fruitName;
	}
	public void setFruitName(String fruitName) {
		this.fruitName = fruitName;
	}
	public String getFruitDesc() {
		return fruitDesc;
	}
	public void setFruitDesc(String fruitDesc) {
		this.fruitDesc = fruitDesc;
	}
	public int getQuantity() {
		return quantity;
	}
	public void setQuantity(int quantity) {
		this.quantity = quantity;
	}
}

To sort it, you may think of Arrays.sort() again, see below example :

package com.mkyong.common.action;

import java.util.Arrays;

public class SortFruitObject{

	public static void main(String args[]){

		Fruit[] fruits = new Fruit[4];

		Fruit pineappale = new Fruit("Pineapple", "Pineapple description",70); 
		Fruit apple = new Fruit("Apple", "Apple description",100); 
		Fruit orange = new Fruit("Orange", "Orange description",80); 
		Fruit banana = new Fruit("Banana", "Banana description",90); 

		fruits[0]=pineappale;
		fruits[1]=apple;
		fruits[2]=orange;
		fruits[3]=banana;

		Arrays.sort(fruits);

		int i=0;
		for(Fruit temp: fruits){
		   System.out.println("fruits " + ++i + " : " + temp.getFruitName() + 
			", Quantity : " + temp.getQuantity());
		}

	}	
}

Nice try, but, what you expect the Arrays.sort() will do? You didn’t even mention what to sort in the Fruit class. So, it will hits the following error :

Exception in thread "main" java.lang.ClassCastException: 
com.mkyong.common.Fruit cannot be cast to java.lang.Comparable
	at java.util.Arrays.mergeSort(Unknown Source)
	at java.util.Arrays.sort(Unknown Source)

To sort an Object by its property, you have to make the Object implement the Comparable interface and override thecompareTo() method. Lets see the new Fruit class again.

public class Fruit implements Comparable<Fruit>{

	private String fruitName;
	private String fruitDesc;
	private int quantity;

	public Fruit(String fruitName, String fruitDesc, int quantity) {
		super();
		this.fruitName = fruitName;
		this.fruitDesc = fruitDesc;
		this.quantity = quantity;
	}

	public String getFruitName() {
		return fruitName;
	}
	public void setFruitName(String fruitName) {
		this.fruitName = fruitName;
	}
	public String getFruitDesc() {
		return fruitDesc;
	}
	public void setFruitDesc(String fruitDesc) {
		this.fruitDesc = fruitDesc;
	}
	public int getQuantity() {
		return quantity;
	}
	public void setQuantity(int quantity) {
		this.quantity = quantity;
	}

	public int compareTo(Fruit compareFruit) {

		int compareQuantity = ((Fruit) compareFruit).getQuantity(); 

		//ascending order
		return this.quantity - compareQuantity;

		//descending order
		//return compareQuantity - this.quantity;

	}	
}

The new Fruit class implemented the Comparable interface, and overrided the compareTo() method to compare its quantity property in ascending order.

The compareTo() method is hard to explain, in integer sorting, just remember

  1. this.quantity – compareQuantity is ascending order.
  2. compareQuantity – this.quantity is descending order.

To understand more about compareTo() method, read this Comparable documentation.

Run it again, now the Fruits array is sort by its quantity in ascending order.

fruits 1 : Pineapple, Quantity : 70
fruits 2 : Orange, Quantity : 80
fruits 3 : Banana, Quantity : 90
fruits 4 : Apple, Quantity : 100

4. Sort an Object with Comparator

How about sorting with Fruit’s “fruitName” or “Quantity”? The Comparable interface is only allow to sort a single property. To sort with multiple properties, you need Comparator. See the new updated Fruit class again :

import java.util.Comparator;

public class Fruit implements Comparable<Fruit>{

	private String fruitName;
	private String fruitDesc;
	private int quantity;

	public Fruit(String fruitName, String fruitDesc, int quantity) {
		super();
		this.fruitName = fruitName;
		this.fruitDesc = fruitDesc;
		this.quantity = quantity;
	}

	public String getFruitName() {
		return fruitName;
	}
	public void setFruitName(String fruitName) {
		this.fruitName = fruitName;
	}
	public String getFruitDesc() {
		return fruitDesc;
	}
	public void setFruitDesc(String fruitDesc) {
		this.fruitDesc = fruitDesc;
	}
	public int getQuantity() {
		return quantity;
	}
	public void setQuantity(int quantity) {
		this.quantity = quantity;
	}

	public int compareTo(Fruit compareFruit) {

		int compareQuantity = ((Fruit) compareFruit).getQuantity(); 

		//ascending order
		return this.quantity - compareQuantity;

		//descending order
		//return compareQuantity - this.quantity;

	}

	public static Comparator<Fruit> FruitNameComparator 
                          = new Comparator<Fruit>() {

	    public int compare(Fruit fruit1, Fruit fruit2) {

	      String fruitName1 = fruit1.getFruitName().toUpperCase();
	      String fruitName2 = fruit2.getFruitName().toUpperCase();

	      //ascending order
	      return fruitName1.compareTo(fruitName2);

	      //descending order
	      //return fruitName2.compareTo(fruitName1);
	    }

	};
}

The Fruit class contains a static FruitNameComparator method to compare the “fruitName”. Now the Fruit object is able to sort with either “quantity” or “fruitName” property. Run it again.

1. Sort Fruit array based on its “fruitName” property in ascending order.

Arrays.sort(fruits, Fruit.FruitNameComparator);

Output

fruits 1 : Apple, Quantity : 100
fruits 2 : Banana, Quantity : 90
fruits 3 : Orange, Quantity : 80
fruits 4 : Pineapple, Quantity : 70

2. Sort Fruit array based on its “quantity” property in ascending order.

Arrays.sort(fruits)

Output

fruits 1 : Pineapple, Quantity : 70
fruits 2 : Orange, Quantity : 80
fruits 3 : Banana, Quantity : 90
fruits 4 : Apple, Quantity : 100
The java.lang.Comparable and java.util.Comparator are powerful but take time to understand and make use of it, may be it’s due to the lacking of detail example.
PS: cited from Cracking the code interview
Write a method to sort an array of strings so that all the anagrams are next to each other.
pg 66
SOLUTION
The basic idea is to implement a normal sorting algorithm where you override the compareTo method to compare the “signature” of each string. In this case, the signature is the alphabetically sorted string.
1 public class AnagramComparator implements Comparator<String> {
2 public String sortChars(String s) {
3 char[] content = s.toCharArray();
4 Arrays.sort(content);
5 return new String(content);
6 }
7
8 public int compare(String s1, String s2) {
9 return sortChars(s1).compareTo(sortChars(s2));
10 }
11 }
Now, just sort the arrays, using this compareTo method instead of the usual one.
12 Arrays.sort(array, new AnagramComparator());

ArrayList vs. LinkedList vs. Vector

http://www.programcreek.com/2013/03/arraylist-vs-linkedlist-vs-vector/

1. List Overview

List, as its name indicates, is an ordered sequence of elements. When we talk about List, it is a good idea to compare it with Set. A Set is a set of unique and unordered elements.

The following is the class hierarchy diagram of Collection. From that you have a general idea of what I’m talking about.

2. ArrayList vs. LinkedList vs. Vector

From the hierarchy diagram, they all implement List interface. They are very similar to use. Their main difference is their implementation which causes different performance for different operations.

ArrayList is implemented as a resizable array. As more elements are added to ArrayList, its size is increased dynamically. It’s elements can be accessed directly by using the get and set methods, since ArrayList is essentially an array. LinkedList is implemented as a double linked list. Its performance on add and remove is better than Arraylist, but worse on get and set methods. Vector is similar with ArrayList, but it is synchronized. ArrayList is a better choice if your program is thread-safe. Vector and ArrayList require space as more elements are added. Vector each time doubles its array size, while ArrayList grow 50% of its size each time.

LinkedList, however, also implements Queue interface which adds more methods than ArrayList and Vector, such as offer(), peek(), poll(), etc.

Note: The default initial capacity of an ArrayList is pretty small. It is a good habit to construct the ArrayList with a higher initial capacity. This can avoid the resizing cost.

3. ArrayList example

ArrayList<Integer> al = new ArrayList<Integer>();
al.add(3);
al.add(2);		
al.add(1);
al.add(4);
al.add(5);
al.add(6);
al.add(6);

Iterator<Integer> iter1 = al.iterator();
while(iter1.hasNext()){
	System.out.println(iter1.next());
}

4. LinkedList example

LinkedList<Integer> ll = new LinkedList<Integer>();
ll.add(3);
ll.add(2);		
ll.add(1);
ll.add(4);
ll.add(5);
ll.add(6);
ll.add(6);

Iterator<Integer> iter2 = ll.iterator();
while(iter2.hasNext()){
	System.out.println(iter2.next());
}

As shown in the examples above, they are similar to use. The real difference is their underlying implementation and their operation complexity.

5. Vector

Vector is almost identical to ArrayList, and the difference is that Vector is synchronized. Because of this, it has an overhead than ArrayList. Normally, most Java programmers use ArrayList instead of Vector because they can synchronize explicitly by themselves.

6. Performance of ArrayList vs. LinkedList

The time complexity comparison is as follows:
arraylist-vs-linkedlist-complexity

* add() in the table refers to add(E e), and remove() refers to remove(int index)

    • ArrayList has O(n) time complexity for arbitrary indices of add/remove, but O(1) for the operation at the end of the list.
    • LinkedList has O(n) time complexity for arbitrary indices of add/remove, but O(1) for operations at end/beginning of the List.

I use the following code to test their performance:

ArrayList<Integer> arrayList = new ArrayList<Integer>();
LinkedList<Integer> linkedList = new LinkedList<Integer>();

// ArrayList add
long startTime = System.nanoTime();

for (int i = 0; i < 100000; i++) {
	arrayList.add(i);
}
long endTime = System.nanoTime();
long duration = endTime - startTime;
System.out.println("ArrayList add:  " + duration);

// LinkedList add
startTime = System.nanoTime();

for (int i = 0; i < 100000; i++) {
	linkedList.add(i);
}
endTime = System.nanoTime();
duration = endTime - startTime;
System.out.println("LinkedList add: " + duration);

// ArrayList get
startTime = System.nanoTime();

for (int i = 0; i < 10000; i++) {
	arrayList.get(i);
}
endTime = System.nanoTime();
duration = endTime - startTime;
System.out.println("ArrayList get:  " + duration);

// LinkedList get
startTime = System.nanoTime();

for (int i = 0; i < 10000; i++) {
	linkedList.get(i);
}
endTime = System.nanoTime();
duration = endTime - startTime;
System.out.println("LinkedList get: " + duration);

// ArrayList remove
startTime = System.nanoTime();

for (int i = 9999; i >=0; i--) {
	arrayList.remove(i);
}
endTime = System.nanoTime();
duration = endTime - startTime;
System.out.println("ArrayList remove:  " + duration);

// LinkedList remove
startTime = System.nanoTime();

for (int i = 9999; i >=0; i--) {
	linkedList.remove(i);
}
endTime = System.nanoTime();
duration = endTime - startTime;
System.out.println("LinkedList remove: " + duration);

And the output is:

ArrayList add:  13265642
LinkedList add: 9550057
ArrayList get:  1543352
LinkedList get: 85085551
ArrayList remove:  199961301
LinkedList remove: 85768810

The difference of their performance is obvious. LinkedList is faster in add and remove, but slower in get. Based on the complexity table and testing results, we can figure out when to use ArrayList or LinkedList. In brief, LinkedList should be preferred if:

  • there are no large number of random access of element
  • there are a large number of add/remove operations