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.

思路：从后往前搜索直到第一个空格字符，注意“a ”的结果是1而不是0，因此需要先排除最后的空格字符。

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        if(s == NULL)   return 0;
        
        int length = strlen(s);
        
        int result = 0;
        for(int i = length - 1; i >= 0 && s[i] == ' '; i--)
            length--;
            
        for(int i = length - 1; i >= 0; i--)
        {
            if(s[i] != ' ')
                result++;
            else
                break;
        }
        
        return result;
    }
};