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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.
Solution1:
偷懒做法，用trim和split。
public class Solution { public int lengthOfLastWord(String s) { String[] res=s.trim().split(" "); return res[res.length-1].length(); } }
Solution2:
两个while，一个检查末尾，一个检查第一个碰到的空格。
public class Solution { public int lengthOfLastWord(String s) { int index=s.length()-1; int count=0; while(index>=0&&s.charAt(index)==‘ ‘) { index--; } while(index>=0&&s.charAt(index)!=‘ ‘) { index--; count++; } return count; } }
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原文地址：http://www.cnblogs.com/Machelsky/p/5871222.html
