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.

解题报告:

    用个变量last保存遍历到当前位置的最后一个空格位置，只要当前字符不是空格，就更新

res，这样下来，最终res里面存放的就是所要求的结果.需要注意的是,last变量的初值.

class Solution {
public:
    int lengthOfLastWord(const char *s) 
    {
        if (s == NULL)
            return 0;
        int res = 0,last = -1;
        for (int i = 0; s[i]; ++i)
        {
            if( s[i] == ' ')
            {
                last = i;
                continue;
            }
            res = i - last;
        }
        return res;
    }
};