LeetCode 58最后一个单词长度

题目描述

给定一个仅包含大小写字母和空格 ’ ’ 的字符串 s,返回其最后一个单词的长度.如果字符串从左向右滚动显示,那么最后一个单词就是最后出现的单词.

如果不存在最后一个单词,请返回 0 .

说明：一个单词是指仅由字母组成、不包含任何空格字符的 最大子字符串.

示例

输入: “Hello World”

输出: 5

题目解析

按照题目的意思,Anyway, to find the length of the last word with letters（Of course it's not up to us whether the spelling is correct or not）.Then start traversing from back to front,统计字符的个数,When a space is encountered, the length of the statistics can be returned.But the string given by the title may give multiple spaces at the end,例如：“hello world ”.In this case, just remove the leading and trailing spaces.There are two ways to remove it：

1. 使用java字符串中的trim方法去掉首尾空格
2. don't show,Skip the last space directly when counting,Counting begins when the first character that is not a space is encountered.

下面是两种实现方式：

1. 对于java,C++,pythonsuch language,There are slice operations for strings,即split.The following is used directlysplit函数,Divide the input string into a string array based on spaces,Return the length of the last string in the array.

       public int lengthOfLastWord(String s) {
         
           String[] split = s.split(" ");
           if (split.length == 0) {
         	//sIf all are spaces, the array after splitting is empty
               return 0;
           }
           return split[split.length - 1].length();
       
       }
   

   程序执行结果：

2. Use the abovesplitThe method is simple and straightforward,But not so fast,The method of counting directly from the back to the front is much faster,代码如下：

   public int lengthOfLastWord(String s) {
         
   
       int count = 0;
       //s = s.trim(); //Remove leading and trailing spaces in advance
       for (int i = s.length() - 1; i >= 0; i--) {
         
           if (s.charAt(i) == ' ' && count == 0) {
         		//Just skip the trailing spaces,没有用trimmethod can do so
               continue;
           }
           if (s.charAt(i) == ' ') {
         	//After skipping all the spaces at the end, if there is a space after it, it proves that the length of the last word has been counted
               break;
           }
           count++;
       }
       return count;
   }