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Leetcode 58. Length of Last Word

时间:2016-08-04 22:49:03      阅读:174      评论:0      收藏:0      [点我收藏+]

58. Length of Last Word

  • Total Accepted: 103816
  • Total Submissions: 345263
  • Difficulty: Easy

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.

 

思路:由后向前遍历,先去掉末尾的空格,然后截取最后一个word。

 

代码:

 1 class Solution {
 2 public:
 3     int lengthOfLastWord(string s) {
 4         int i=s.size()-1,j;
 5         while(i>=0&&s[i]== ) i--;
 6         j=i;
 7         while(i>=0&&s[i]!= ) i--;
 8         return j-i;
 9     }
10 };

 

Leetcode 58. Length of Last Word

原文:http://www.cnblogs.com/Deribs4/p/5738411.html

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