Length of last word

Question

Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.

If the last word does not exist, return 0.

Note: A word is defined as a character sequence consists of non-space characters only.

For example:

s               result 

"Hello World"    5
"a "             1
" a "            1  
"  ba "          2  
"ba    "         2

The following is my code:

int lengthOfLastWord(const char* s)
{                   
    const char* end=s;
    while (*end != '\0')
    {
        ++end;
    }
    --end;

    while ((end >= s) && (*end == ' '))
    {
        --end;
    }

    const char* start = end;
    while ((start >= s) && (*start != ' '))
    {
        --start;
    }

    return end-start;
}

Answer 1

This is just the @palacsint solution extended to ignore trailing spaces.

int lengthOfLastWord2(const char* input)
{
    int result = 0;
    int last_result = 0;

    while (*input != '\0') {
        if (*input != ' ') {
            result++;
        } else if (result) {
            last_result = result;
            result = 0;
        }
        input++;
    }
    return result ? result : last_result;
}

Answer 2

(Just a quick note, I don't have too much time now.) Here is another approach:

int lengthOfLastWord2(const char* input)
{                   
    int result = 0;
    while (*input != '\0') {
        if (*input != ' ') {
            result++;
        } else {
            result = 0;
        }
        input++;
    }

    return result;
}

Please note that it returns zero when the last character is a space (it was not specified in the question).

Answer 3

All of the other answers assume "space" is only ' ', when in fact it can be \f, \n, \t \r or \v (see man isspace). The original question states "empty space characters" which implies there is more than one meaning (to me) that it is including all space characters. Some of the examples previous included would fail

"joseph\n"
"joseph\v"

counting this as 7 characters instead of 6.

So I thought I would rewrite this using the isspace standard library function, and test it against these other cases as well.

#include <stdio.h>
#include <string.h>

int lengthOfLastWord3(const char *input )
{
  const char *end = input;
  const char *last_word_start = NULL;
  const char *last_word_end = NULL;
  char prev_char = '\0';
  int word_length = 0;

  while ( *end != '\0')
    {
      if ( !isspace( *end ) && 
           ( isspace( prev_char ) || ( prev_char == '\0' )))
        {
          last_word_start = end;
          last_word_end = end+1;
        }
      else if ( !isspace( prev_char ) && ( isspace( *end ) ) )
        {
          last_word_end = end;
        }
      else if ( !isspace( prev_char ) && ( !isspace( *end ) ) )
        {
          last_word_end = end+1;
        }

      prev_char = *end;

      end++;
    }

  if ( last_word_start )
    {
      word_length = last_word_end - last_word_start;
    }

  return( word_length );
}

This works for a bunch of additional cases, including entries where \n \t \v \f are used.