Nolan Lefler

Thanks for the suggestions. I haven't encountered HashMaps yet, but I'll look into them.

But I've never heard the main method should only being one line of code. That's quite new to me.

Mr Tuah wrote:How about this? it will show all words that found in sentences

That looks nice to me Mr Tuah, but my code requires the number of times a word occurs in a string to be counted.

Ryan Sykes wrote:I'll let someone more experienced comment on elegance, but I wanted to point out that your code fails to identify words that are adjacent to punctuation marks. The presence of a period, comma or question mark next to a word will give you faulty results.

I've fixed that issue. However, if somebody enters a word that has punctuation in the middle of the word, such as 'it's' or 'jack-in-the-box' it will make these words 'its' and 'jackinthebox'

Can you tell me how elegant this solution is for the problem? I tried using as little code as required. This is basically listing how many times a word appears in a line of text. I have converted the text to a string array.

import java.io.*; class ListCheck { public static void main(String[] args) throws IOException { InputStreamReader converter = new InputStreamReader(System.in); BufferedReader input = new BufferedReader(converter); System.out.print("\nPlease enter some text: "); String textInput = input.readLine(); int counter; if(textInput == null || textInput.length() == 0) System.out.print("\nYou have entered no text.\n"); else { // Split each word of text into an array. Seperate words by spaces. String text[] = textInput.split(" +"); // Remove punctuation from words. for (int i = 0; i < text.length; i++) { text[i] = text[i].replaceAll("\\W", ""); } // Cycle through array, looking for duplicate instances of words. for(int i = 0; i < text.length; i++) { // Check to see if the word we are looking at appears previously in the array. If so, don't continue looking. if(!AlreadyInList(text,i)) { counter = 0; for(int j = i; j < text.length; j++) { if(text[i].equals(text[j])) counter++; } System.out.print("\n\"" + text[i] + "\" appears in the text " + counter + " time(s)."); } } System.out.println(); } } public static boolean AlreadyInList(String [] text, int index) { boolean inList = false; // No need to check first word in the array. We know it is unique. if(index != 0) { for (int i = 0; i < index; i++) { if(text[i].equals(text[index])) { inList = true; break; } } } return inList; } }

[deleted]

I have just started with HF Java, and this is the code I wrote. It works for me.

public class BeerSong { public static void main(String[] args) { int beerNum = 99; String word = "bottles"; while (beerNum > 0) { System.out.println(beerNum + " " + word + " of beer on the wall"); System.out.println(beerNum + " " + word + " of beer."); System.out.println("Take one down."); System.out.println("Pass it around."); beerNum = beerNum - 1; if (beerNum == 1) { word = "bottle"; } if (beerNum > 0) { System.out.println(beerNum + " " + word + " of beer on the wall"); } else { System.out.println("No more bottles of beer on the wall"); } } } }

Hi Alexander,

Welcome to the wonderful world of Android!

I would suggest getting the updated software for Eclipse, the Android SDK, and also the ADT plug-in. These are required tools. The following link provides you everything you need to know:

http://developer.android.com/sdk/installing.html#InstallingADT

As for being restricted to a specific platform, you can use whichever platform you like as long as you download the required components.

Enjoy!
Jay