题目：

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：

首先，统计字符串总长度；其次，找到倒数的第一个不为空格的字符；最后找到字符前的第一个空格。

代码：

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = 0;
        while(s[len] != '\0')
        {
            len++;
        }
        int i = len - 1;
        int j = len - 1;
        for(; i >= 0; i--)
        {
            if(s[i] != ' ')
            {
                j = i;
                break;
            }
        }
        
        if(i == -1)
        {
            return 0;
        }
        for(; i >= 0; i--)
        {
            if(s[i] == ' ')
            {
                break;
            }
        }
        
        if(i == -1)
        {
            return j + 1;
        }
        else
        {
            return j - i;
        }
    }
};