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.

C++代码实现：

#include<iostream>
#include<vector>
#include<string>
#include<sstream>
#include<cstring>
using namespace std;

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        istringstream istr(s);
        vector<string> vec;
        string ss;
        while(istr>>ss)
            vec.push_back(ss);
        if(vec.size()==0)
            return 0;
        return vec[vec.size()-1].length();
    }
};

int main()
{
    const char *s = "Hello World";
    Solution ss;
    cout<<ss.lengthOfLastWord(s)<<endl;
}

本来是很简单的一个题，但是因为没有判断字符串全部由空格组成。这时经过istringstream处理之后压入到vector中的元素将是0个，因此vec.size()将是0，所以最后一个将返回运行时错误。（细节决定成败啊）

 以前怎么想到那么麻烦的方法呢，明明用双指针就可以搞定的事啊。。

class Solution {
public:
    int lengthOfLastWord(const char *s) {
        if(s==NULL)
            return 0;
        int len=0;
        const char *p=s;
        const char *q=NULL;
        while(*p!='') ++p;
        p--;
        while(p>=s&&*p==' ') --p;
        if(p<s)
            return 0;
        q=p;
        while(q>=s&&*q!=' ') q--;
        len=p-q;
        return len;
    }
};