[LeetCode]58、Length of Last Word

题目描述：

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.

Example:

Input: "Hello World"
Output: 5

思路：

给一个字符串，判定最后一个单词的字母的个数。可以从后往前计数，直到遇见空格为止。
预处理：先把字符串后面的空格都去掉：
先用length表示s的长度-1，判定这个s.charAt(length)是否等于空格，即最后一位是否为空格，如果是length--，然后继续判断，直到不是空格开始计数。
从i=length开始，i递减遍历s，如果等于空格则break，如果不等则count++
返回count即可

 1 public class Solution58 {
 2     public int lengthOfLastWord(String s){
 3         int count = 0;
 4         if(s.isEmpty()) return 0;
 5         int length = s.length()-1;
 6         while(length>0 && s.charAt(length) ==' ') --length;
 7         for(int i = length;i>=0;i--){
 8             if(s.charAt(i) == ' '){
 9                 break;
10             }else {
11                 count++;
12             }
13         }
14         //if(count == s.length()){return 0;}
15         return count;
16     }
17     public static void main(String[] args) {
18         // TODO Auto-generated method stub
19         String s = "Hello world";
20         Solution58 solution58 = new Solution58();
21         System.out.println(solution58.lengthOfLastWord(s));
22     }
23 
24 }

 

来源：https://www.cnblogs.com/zlz099/p/8144964.html