[LeetCode] 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.

这题看似很简单，本来想睡前切一道怡情，结果切到一点钟，东改西改还是没AC....

因为edge case 很多，考虑一不周全就是错的，比如 "", " ", " a", " a", "a     b     " 这几个case。

最后我AC的思路是用 i 来记录当前的词长，如果遇到空格则说明此词完毕，把 j = i; i = 0 然后继续直到串尾巴'\0'

最后判断 i 是否为0，如果是则返回j，否则返回i

 int lengthOfLastWord(const char *s) {
     int i=, j=;
     while (*s != '\0') {
         if (*s == ' ') {
             if (i != ){
                 j = i;
                 i = ;
             }
         }
         else {
             i++;
         }
         s++;
     }
     if (i ==  ) return j;
     return i;
 }

囧