[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''

最后判断 i 是否为0，如果是则返回j，否则返回i

 1 int lengthOfLastWord(const char *s) {
 2     int i=0, j=0;
 3     while (*s != '') {
 4         if (*s == ' ') {
 5             if (i != 0){
 6                 j = i;
 7                 i = 0;
 8             }
 9         }
10         else {
11             i++;
12         }
13         s++;
14     }
15     if (i == 0 ) return j;
16     return i;
17 }

囧