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.

提示：

此题比较简单，因为输入的字符串只会包含字母和空格，所以只要从右往左扫描即可。

代码：

class Solution {
public:
    int lengthOfLastWord(string s) {
        if (s.size() == 0) return 0;
        int idx = s.find_last_not_of(‘ ‘), i;
        if (idx == - 1){
            return 0;
        } else {
            for (i = idx; i > -1; --i) {
                if (s[i] == ‘ ‘) return idx - i;
            }
        }
        return ++idx;
    }
};

【LeetCode】58. Length of Last Word

原文：http://www.cnblogs.com/jdneo/p/4766699.html