[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.

分析

无

代码

/**————————————* 日期：2015-02-06* 作者：SJF0115* 题目: 58.Length of Last Word* 网址：https://oj.leetcode.com/problems/length-of-last-word/* 结果：AC* 来源：LeetCode* 博客：—————————————**/;class Solution {public:int lengthOfLastWord(const char *s) {int len = strlen(s);int i = len – 1;int lastLen = 0;// 去掉空格while(s[i] == ‘ ‘){–i;}//whilewhile(i >= 0 && s[i] != ‘ ‘){++lastLen;–i;}//whilereturn lastLen;}};int main(){Solution s;char *str = ” q f “;//”hello world”;int result = s.lengthOfLastWord(str);// 输出cout<<result<<endl;return 0;}

运行时间

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