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LeetCode58 Length of Last Word

时间:2016-09-20 00:00:24      阅读:321      评论:0      收藏:0      [点我收藏+]

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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. (Easy)

分析:

注意把最后的空格处理掉即可。

代码:

 1 class Solution {
 2 public:
 3     int lengthOfLastWord(string s) {
 4         int start = s.size() - 1;
 5         int result = 0;
 6         while (s[start] ==  ) {
 7             start--;
 8         }
 9         for (int i = start; i >= 0; --i) {
10             if (s[i] !=  ) {
11                 result++;
12             }
13             else {
14                 return result;
15             }
16         }
17         return result;
18     }
19 };

 

LeetCode58 Length of Last Word

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原文地址:http://www.cnblogs.com/wangxiaobao/p/5886920.html

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