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.

解法一

分三步：

1. 统计字符串的有效长度；
2. 滤掉最后的空格；
3. 统计最后一个单词的长度。

代码如下：

C++ code
    int lengthOfLastWord(const char *s) {
        int i, j, len = 0;
        for (i = 0; s[i] != ‘\0‘; ++i) ;
        for (  --i; s[i] ==  ‘ ‘; --i) ;
        for (j = i; j >= 0 && s[j] != ‘ ‘; --j, ++len) ;
        return len;
    }

解法二

一趟找到最后那个单词的长度。

C++ code
    int lengthOfLastWord(const char *s) {
        int len = 0, last = 0;
        while (*s)
            if (*s++ == ‘ ‘) {
                last = len > 0 ? len : last;
                len = 0;
            }
            else ++len;

        return len > 0 ? len : last;
    }

LeetCode[string]: Length of Last Word