Leet Code OJ 58. Length of Last Word [Difficulty: Easy]

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.

翻译：
给定一个字符串，它是由大小写字母和空格组成的，返回字符串中最后一个单词的长度。
如果最后一个单词不存在，就返回0.
提示：一个单词是指一个只由非空格的字符组成的字符串序列。

分析:
有一点题目中没有说明的是，假如最后是空格的话，如何处理。经过测试，发现最后是空格的话，还是返回上一个单词的长度，并不会清零。

代码：

public class Solution {public int lengthOfLastWord(String s) {char[] arr=s.toCharArray();int count=0;boolean needClean=false;for(int i=0;i<arr.length;i++){if(arr[i]!=' '){if(needClean){count=0;needClean=false;}count++;}else{needClean=true;}}return count;}
}

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