identifying indices(java)

Question

it makes me sick ..can you help me with this one? my problem is to identify the whitespaces on my java program and the indices of it but i dont know how to identify the indices(JAVA). heres my code:

import java.util.*;

public class CountSpaces
{
public static void main (String[] args)
  {
   System.out.print ("Enter a sentence or phrase: ");
   Scanner input=new Scanner(System.in);
   String str=input.nextLine();
   int count = 0;
   int limit = str.length();
    for(int i = 0; i < limit; ++i)
    {
     if(Character.isWhitespace(str.charAt(i)))
     {
      ++count;
     }
    }

thanks ahead.

Brian Roach · Answer 1 · 2011-09-06T07:14:21.783

4

Use an ArrayList to record the indices. This also eliminates the need for a count as the number of entries in the list is the number of occurrences.

ArrayList<Integer> whitespaceLocations = new ArrayList<Integer>();
for(int i = 0; i < limit; ++i)
{
    if(Character.isWhitespace(str.charAt(i)))
    {
        whitespaceLocations.add(i);
    }
}

System.out.println("Whitespace count: " + whitespaceLocations.size());
System.out.print("Whitespace is located at indices: ");
for (Integer i : whitespaceLocations)
{
    System.out.print(i + " "); 
}

System.out.println();

edited Sep 06 '11 at 07:14

answered Sep 06 '11 at 06:49

Brian Roach

76,169
12
136
161

Why you are using `i.toString()` in `System.out.print(i.toString() + " ");`? This will work `System.out.print(i + " ");`. **OR** Simple and mostly used [in my opinion] `System.out.print(i);`. Also *no need* of `System.out.println();` at the end. – Harry Joy Sep 06 '11 at 07:00
While you're correct in that the former isn't needed ... it's going to happen anyway ::shrug::. The latter on the other hand, is for readability. Your version would print something akin to `Whitespace is located at indices: 251022` with no carriage return whereas the code above prints `Whitespace is located at indices: 2 5 10 22`. – Brian Roach Sep 06 '11 at 07:07
Oh! yes. Didn't noticed that you are using `print` and not `println`. But still thinks you should remove toString(); it is unnecessary. ;-) – Harry Joy Sep 06 '11 at 07:09
Pedantic point - I would favour LinkedList in this case as the list has to dynamically resize and there is no need for efficient random access. I would also declare it as List rather than explicitly specifying the list type. – Adamski Sep 06 '11 at 07:32
Yeah .. I thought of that after posting, but decided to leave it ::shrug:: In this case I don't think it's a huge deal. I could argue that in a real word scenario you might want to know where the third space is. – Brian Roach Sep 06 '11 at 07:38

score 2 · Answer 2 · answered Sep 06 '11 at 06:41

if(Character.isWhitespace(str.charAt(i)))

You already doing the most of it. If the above condition is true, then at index i you have the whitespace character. How ever, if you need to keep track of all the indices, then copy the index i to an array in the if.

identifying indices(java)

2 Answers2