[leetcode] 18. Length of Last Word

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。