Leetcode: Length of Last Word

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.

可能会忽略末尾为“ ”的情况

法一：

1 public class Solution {
2     public int lengthOfLastWord(String s) {
3         if (s==null || s.length()==0) return 0;
4         s = s.trim();
5         String[] array = s.split(" ");
6         String res = array[array.length-1];
7         return res.length();
8     }
9 }

如果没有第4行，遇到s == " "的情况， s.split()之后会出错

 

Split 函数Trailing empty strings are not included in the resulting array. The string "boo:and:foo", for example, yields the following results with these expressions:

Regex	Result
:	{ "boo", "and", "foo" }
o	{ "b", "", ":and:f" }

 

 

法二（推荐方法）：

1 public class Solution {
2     public int lengthOfLastWord(String s) {
3         s = s.trim();
4         int index = s.length()-1;
5         while (index>=0 && s.charAt(index)!=' ') 
6             index--;
7         return (s.length()-index-1);        
8     }
9 }