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.

分析：

简单模拟思想。去掉尾部的空格并统计是字符的个数直到再次遇到空格就停止统计

class Solution {

public:

    int lengthOfLastWord(string s) {

        int strLen=s.size();

        int len=0;

        int i=0;

        while(s[strLen-i-1]==' ')

            i++;

        for(;i<strLen;i++)

        {

            if(s[strLen-i-1]>='A'&&s[strLen-i-1]<='Z' || s[strLen-i-1]>='a'&&s[strLen-i-1]<='z' )

                len++;

            else

                break;

        }

        return len;

    }

};

或者用STL来写：

class Solution {

public:

    int lengthOfLastWord(string s) {

        int i = s.find_last_not_of(' ');

        return (i == string::npos) ? 0 : (i - s.find_last_of(' ', i));

    }

};

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原文地址：http://blog.csdn.net/ebowtang/article/details/50498956

原作者博客：http://blog.csdn.net/ebowtang