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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.
代码：
class Solution { public: int lengthOfLastWord(string s) { for ( int i = s.length()-1; i >=0; --i ) { if (s[i]==‘ ‘) { s.erase(s.end()-1); } else { break; } } const size_t len = s.length(); int ret = 0; for ( size_t i = 0; i < len; ++i ) { if (s[i]!=‘ ‘) { ++ret; continue; } if (s[i]==‘ ‘) { ret = 0; continue; } } return ret; } };
tips：
先把后面的空格都去掉。然后从头遍历，最后留下的ret就是最后一个单词的长度。
还有STL的一个做法，明天再看。
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原文地址：http://www.cnblogs.com/xbf9xbf/p/4496016.html
