[LeetCode] Length of last word

Length of last word

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, 
Given s = "Hello World",
return 5.

这是一道实现题，从后往前扫描，如何判断最后一个word,有以下几种情况

1.“word   ”

2. "       "

3. "word"

4."  word  "

int lengthOfLastWord(const char *s) {
    int length = strlen(s);
    int count = 0;
    for(int i= length-1; i>=0;i--){
        if(s[i] == ' '){
            if(count == 0) continue;
            else return count;
        }
        count++;
    }
    return count;
}

Java解法：

利用split()函数，通过空格“ ”将原字符串分组

public int lengthOfLastWord(String s) {
        String [] splitStrings = s.split(" ");
        	for(int i=splitStrings.length-1;i>=0;i--){
        		if(splitStrings[i].length()>0)
        			return splitStrings[i].length();
        	}
        	return 0;
    }

• public String[] split(String regex)

  Splits this string around matches of the given regular expression.

  This method works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array.

  The string "boo:and:foo", for example, yields the following results with these expressions:

  Regex	Result
  :	{ "boo", "and", "foo" }
  o	{ "b", "", ":and:f" }

  Parameters:

  regex - the delimiting regular expression

  Returns:

  the array of strings computed by splitting this string around matches of the given regular expression

  Throws:

  PatternSyntaxException - if the regular expression's syntax is invalid