LeetCode(58):Length of Last Word

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.

题意:给定一个字符串,判断最后一个不包括“ ”字符串的长度。

思路:首先如果字符串如果为空,则返回0,否则,从字符串的最后一个字符开始。查找第一个不为“ ”的索引,然后开始计数,直到找到下一个“ ”返回计数器的值。

代码:

public int lengthOfLastWord(String s) {
        int count = 0;
         if(s.length()<=0) return 0;
         int index = s.length() - 1 ;
         while(index>=0&&s.charAt(index)==' ')
         {
             index--;
         }
         while(index>=0)
         {
             if(s.charAt(index)!=' ')
             {
                 count++;
                 index--;
                 if(index<0) return count;
             }else 
             {
                 return count;
            }
         }
         return 0;
    }
原文地址:https://www.cnblogs.com/Lewisr/p/5117833.html