java comparator, comparing strings by number of words

I need some help with java comparator. I need to compare strings by the number of words they contain.

For exampl开发者_Python百科e, "hello" comes before "I see" which comes before "I see you".

Anyone got any ideas how I would do this? Thank you in advance for any help you could give.

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import java.util.Comparator;
import java.util.StringTokenizer;

public class WordCountComparator implements Comparator<String> {
    public int compare(String str1, String str2) {
        if (str1 == str2 || (str1 == null && str2 == null)) {
            return 0;
        } else if (str1 == null) {
            return -1;
        } else if (str2 == null) {
            return 1;
        }

        int len1 = new StringTokenizer(str1, " \t\r\n").countTokens();
        int len2 = new StringTokenizer(str2, " \t\r\n").countTokens();
        return len1 < len2 ? -1 : len1 == len2 ? 0 : 1;
    }
}

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Write a comparator for a String type. In the compareTo method, simply call string.split(" ") to break it down to 'words' (if that's sufficient), then return the comparison between those two integers.

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string.split(" ") will return an array, where each element is a word. So comparing the number of elements in the array would give you a rough implementation of this. Obviously you may need to handle line-breaks, tabs etc., so keep that in mind.

Apache commons also has some useful methods to do this.

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 public class StringComparator implements Comparator<String> {

      public int compare(String o1, String o2) {
        if(o1==null && o2 ==null)
        return 0;
        String[] bo1 = o1.split("\\s");
        String[] bo2 = o2.split("\\s");
        return bo1.length >= bo2.length ? bo1.length > bo2.length ? 1 : 0 : -1;
      }
}