标签:leetcode
前段时间忙平台访问分析的统计库开发,有一个多月没写java了。。找找手感,刷刷题吧。。
去leetcode上挑了个简单的,题目是这样的。
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
.
解决方案如下:
public class Solution { public int lengthOfLastWord(String s) { String[] arr=s.split(" "); if(arr.length==0){ return 0; } return arr[arr.length-1].length(); } }将给定字符串s根据空格进行分隔,共有两种情况:
1、s是一堆空格构成的“空串”,调用split方法会返回一个长度为0的空数组。
2、s中有word,分隔之后取最后一个即可~
标签:leetcode
原文地址:http://blog.csdn.net/exceptional_derek/article/details/43342693