54：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 consist

解题代码一（从后往前扫描）：

// 逆序扫描
// 时间复杂度 O(n)，空间复杂度 O(1)
class Solution {
public:
int lengthOfLastWord(string s) {
int len = 0;

int i = s.size() - 1;
while (i >= 0)
if (s[i] == ' ') --i;
else break;

while (i >= 0)
if (s[i] != ' ') ++len, --i;
else break;
return len;
}
};

解题代码二（从前往后扫描）：

// 顺序扫描，记录每个 word 的长度
// 时间复杂度 O(n)，空间复杂度 O(1)
class Solution {
public:
int lengthOfLastWord(const string& s) {
int len = 0;
for (int i = 0; i < s.size(); ) {
if (s[i] != ' ') ++len, ++i;
else {
++i;
if (i < s.size() && s[i] != ' ') ++len;
}
}
return len;
}
};