leetcode 58. Length of Last Word

/*
leetcode 58. 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.
对于一句话找出最后一个单词的长度
解题思路：采用一个计数的变量count，从后面找，一旦找到一个为字母，count+1,
如果后面的字符不是字母，且count>0,就可以输出。
*/

#include <iostream>
#include <vector>
#include <string>

using namespace std;

class Solution {
private:
//判断是否为字母
bool IsAlpha(char c)
{
return (c >= 'a' && c <= 'z') || (c >= 'A' && c <= 'Z');
}
public:
int lengthOfLastWord(string s)
{
//从后面开始
int i = s.length() - 1;
int count = 0;
while (i >= 0)
{
if (IsAlpha(s[i]))
{
++count;
--i;
}
else
{
if (count)
return count;
--i;
}
}
return count;
}
};

void TEST()
{
Solution sol;
string s1("this is a test!");
string s2("this is a !");
string s3("this is!");
string s4("Hello World");
cout << sol.lengthOfLastWord(s1) << endl;
cout << sol.lengthOfLastWord(s2) << endl;
cout << sol.lengthOfLastWord(s3) << endl;
cout << sol.lengthOfLastWord(s4) << endl;
}

int main()
{
TEST();

return 0;
}