这个题目很简单，给一个字符串，然后返回最后一个单词的长度就行。题目如下：

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.

我的思路就是先把字符串反转，然后找到第一个‘ ‘后弹出，这个需要注意的大概就是很可能最后不是字符，而是一个标点，所以要判断一下。题解如下：

class Solution {public:	int lengthOfLastWord(const char *s) 	{		string tmp = s;		reverse(tmp.begin(), tmp.end());		int sum, i, len;		sum = i = 0;		len = tmp.length();		while (i < len)		{			if (isalpha(tmp[i]))			{				sum++;				i++;				break;			}			else			{				i++;			}		}		while (tmp[i] != ‘ ‘ && i < len)		{			sum++;			i++;		}		return sum;	}};

如上，不难。

[leetcode] 18. Length of Last Word