问题描述:
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
.
原问题链接:https://leetcode.com/problems/length-of-last-word/
问题分析
这个问题相对比较简单,只是需要注意几个小的细节。一个是当给定的String是空的时候,需要判断并返回对应的结果。因为这里定义的word是由非空的字符组成,所以首先从后往前找第一个非空的字符。如果没有找到,则返回0。否则就从这个点开始一直往前判断到当前有空的字符或者到字符串的开头。然后再将两个位置相减就得到最后的word长度。
代码实现如下:
public class Solution { public int lengthOfLastWord(String s) { if(s == null || s.length() < 1) return 0; int start, end; for(end = s.length() - 1; end >= 0 && s.charAt(end) == ' '; end--); if(end < 0) return 0; for(start = end; start >= 0 && s.charAt(start) != ' '; start--); return end - start; } }