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.
Notice
A word is defined as a character sequence consists of non-space characters only.
Given s = "Hello World", return 5.
LeetCode上的原题，请参见我之前的博客Length of Last Word。
解法一：
class Solution { public: /** * @param s A string * @return the length of last word */ int lengthOfLastWord(string& s) { if (s.empty()) return 0; int res = 0; if (s[0] != ' ') res = 1; for (int i = 1; i < s.size(); ++i) { if (s[i] != ' ') { if (s[i - 1] == ' ') res = 1; else ++res; } } return res; } };
解法二：
class Solution { public: /** * @param s A string * @return the length of last word */ int lengthOfLastWord(string& s) { int tail = s.size() - 1, res = 0; while (tail >= 0 && s[tail] == ' ') --tail; while (tail >= 0 && s[tail] != ' ' ) { --tail; ++res; } return res; } };
本文转自博客园Grandyang的博客，原文链接：求末尾单词的长度[LintCode] Length of Last Word ，如需转载请自行联系原博主。