[leetcode] 58.Length of Last Word

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.

题意：

给定一个字符串，包含大小写或者空格。

思路：

把字符串先反转，然后找第一个word的长度。

以上。

代码如下：

class Solution {
public:
int lengthOfLastWord(string s) {
if(s.empty())return 0;
int first = 0, second = s.length() - 1;
while(first < second) {
swap(s[first], s[second]);
first++;
second--;
}
first = 0;
while(first < s.size() && s[first] == ' ')first++;
second = first;
while(second < s.size() && s[second] != ' ')second++;
return second - first;
}
};