58. Length of Last Word(easy, 字符串问题)

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.

我的方法是：

1. 先消除字符串首、尾的空格;
2. 再从字符串尾向首查找，直到找到空格或者指针到队首都没找到空格为止，返回最后一个 word 的长度.

我方法，我代码：
(O(n)) time, (O(1)) extra space.

// 解题关键：消除 s 的首尾空格
int lengthOfLastWord(string s) {
	const int n = s.size() - 1;
	if (s.size() == 0) return 0;

	int begin = 0, end = n;

	//消除字符串首尾空格
	if (s[0] == ' ') {
		for (int i = 0; i < s.size(); i++) {
			if (s[i] != ' ') {
				begin = i;
				break;
			}
		}
	}

	if (s[n] == ' ') {
		for (int i = n; i >= 0; i--) {
			if (s[i] != ' ') {
				end = i;
				break;
			}
		}
	}

	// 从尾部开始查找
	for (int i = end; i >= begin; i--) {
		if (s[i] == ' ') return end - i;
		else if (i == begin) return end - begin + 1;
	}
}