首页 > 其他 > 详细

58. Length of Last Word

时间:2018-12-23 10:39:40      阅读:143      评论:0      收藏:0      [点我收藏+]

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.

Example:

Input: "Hello World"
Output: 5

solution1:
class Solution {
    public int lengthOfLastWord(String s) {
        int res = 0;
        if(s.length()==0||s==null){
            res = 0;
        }
        else{
            int i = s.length()-1;
           while(i>=0&&s.charAt(i)==‘ ‘)
               i--;
            while(i>=0&&s.charAt(i)!=‘ ‘){
                i--;
                    res++;
            }
        }
        return res;
    }
}

就是先把尾部的空字符去掉,然后从尾部开始计算长度。

solution2:

熟用method,暴力一行解法

public int lengthOfLastWord(String s) {
    return s.trim().length()-s.trim().lastIndexOf(" ")-1;
}

  

 

58. Length of Last Word

原文:https://www.cnblogs.com/wentiliangkaihua/p/10163224.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!