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recognize upper & lower cases

is there a way to count the number of upper and lower case entered in a string..having in mind that the user might enter other than letters..like spaces and other characters..
plus i have to add all the characters in that string..
0
ms_lost
Asked:
ms_lost
2 Solutions
 
objectsCommented:
The Character class has various methods for checking character types, including isLowerCase() and isUpperCase().
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ms_lostAuthor Commented:
yeah but how to check that.. since the user will enter a string mix of both
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objectsCommented:
Just loop thru the string checking each character:

int lc = 0;
for (int i=0; i<s.length(); i++)
{
  if (Character.isLowerCase(s.charAt(i))) lc++;
}
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Jim CakalicSenior Developer/ArchitectCommented:
That will work, although a more performant solution would get the char[] of the String instead of using charAt:

    char[] chars = str.toCharArray();
    for (int i = chars.length - 1; i >= 0; --i) {
        ++numChars;
        if (Character.isLowerCase(chars[i])) {
            ++numLower;
        } else if (Character.isUpperCase(chars[i])) {
            ++numUpper;
        }
    }

Jim
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objectsCommented:
Sorry, I was just focussing on the question and how to count lower case character, and was not convcerned with the performance. And as the string was being entered by the user, and as such would generally be short, then the performance difference would be minimal anyway.
If you really wanted a fast solution that would handle long strings then looping thru the string is not the answer anyway :)
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objectsCommented:
If performance is an issue then the following is a better solution:

char[] c = Arrays.sort((s+"azAZ").toCharArray());
int nlower = Arrays.binarySearch(c, 'z') -
   Arrays.binarySearch(c, 'a') - 1;
int nupper = Arrays.binarySearch(c, 'Z') -
   Arrays.binarySearch(c, 'A') - 1;
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Jim CakalicSenior Developer/ArchitectCommented:
Hmm. Didn't mean to sound as if I was criticizing -- just extending your remarks. (At least I didn't post as an answer, eh?)

Jim
:-)
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objectsCommented:
No offence taken :-)
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girionisCommented:
No comment has been added lately, so it's time to clean up this TA.

I will leave a recommendation in the Cleanup topic area that this question is:

- split points between objects@idg and jim_cakalic@idg

Please leave any comments here within the
next seven days.

PLEASE DO NOT ACCEPT THIS COMMENT AS AN ANSWER !

girionis
Cleanup Volunteer
0
 
Jim CakalicSenior Developer/ArchitectCommented:
Friday is _cleanup_ day! OK by me. :-)
Jim
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