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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.
思路：
从后面开始找
package string; public class LengthOfLastWord { public int lengthOfLastWord(String s) { int n = 0; if (s == null || (n = s.length()) == 0) return 0; int postSpaceIndex = -1; for (int i = n - 1; i >= 0; --i) { if (s.charAt(i) != ‘ ‘) { postSpaceIndex = i; break; } } int count = 0; for (int i = postSpaceIndex; i >= 0; --i) { if (s.charAt(i) != ‘ ‘) ++count; else break; } return count; } public static void main(String[] args) { // TODO Auto-generated method stub String s = "Hello world "; LengthOfLastWord l = new LengthOfLastWord(); System.out.println(l.lengthOfLastWord(s)); } }
LeetCode - Length of Last Word
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原文地址：http://www.cnblogs.com/shuaiwhu/p/5079835.html
