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.
Example:
Input: "Hello World" Output: 5
找出字符串中最后一个单词的长度。
利用istringstream来去除字符串中的空格，保存最后一个单词并返回它的长度即可。
class Solution { public: int lengthOfLastWord(string s) { string word; istringstream in(s); while (in >> word) {} return word.size(); } }; // 6 ms
使用遍历判断空格。
如果最后一位非‘ ’，则累计字母，如果如要空格并且计数非0跳出循环。
class Solution { public: int lengthOfLastWord(string s) { int len = 0; for (int i = s.size() - 1; i >= 0; i--) { if (s[i] == ‘ ‘) { if (len != 0) break; } else len++; } return len; } }; // 6 ms
[LeetCode] Length of Last Word
原文：http://www.cnblogs.com/immjc/p/7843681.html