C ++函数来计算字符串中的所有单词 [英] C++ function to count all the words in a string

问题描述

在一个面试中我被问过这个问题，显然这是一个简单的问题，但是对我来说还不是很明显。所有的字。如果他们重复无关紧要。只是总计数像一个文本文件字数。单词是由空格分隔的任何东西，标点符号无关紧要，只要它是单词的一部分。

例如：
一个非常，非常，非常非常大的狗吃了我的家庭作业！ ==> 11个字

我的算法只是通过寻找空格和增加计数器直到我命中null。因为我没有得到的工作，被要求离开后，我想我的解决方案不好？任何人都有更聪明的解决方案？我缺少一些东西？

解决方案

一个不那么聪明，更明显的程序员-team方法。

  #include< cctype& 
 
 int CountWords（const char * str）
 {
 if（str == NULL）
 return error_condition; //让需求定义这个... 
 
 bool inSpaces = true; 
 int numWords = 0; 
 
 while（* str！= NULL）
 {
 if（std :: isspace（* str））
 {
 inSpaces = true; 
} 
 else if（inSpaces）
 {
 numWords ++; 
 inSpaces = false; 
} 
 
 ++ str 
} 
 
 return numWords; 
} 
  

I was asked this during an interview and apparently it's an easy question but it wasn't and still isn't obvious to me.

Given a string, count all the words in it. Doesn't matter if they are repeated. Just the total count like in a text files word count. Words are anything separated by a space and punctuation doesn't matter, as long as it's part of a word.

For example: A very, very, very, very, very big dog ate my homework!!!! ==> 11 words

My "algorithm" just goes through looking for spaces and incrementing a counter until I hit a null. Since i didn't get the job and was asked to leave after that I guess My solution wasn't good? Anyone have a more clever solution? Am I missing something?

解决方案

A less clever, more obvious-to-all-of-the-programmers-on-your-team method of doing it.

#include <cctype>

int CountWords(const char* str)
{
   if (str == NULL)
      return error_condition;  // let the requirements define this...

   bool inSpaces = true;
   int numWords = 0;

   while (*str != NULL)
   {
      if (std::isspace(*str))
      {
         inSpaces = true;
      }
      else if (inSpaces)
      {
         numWords++;
         inSpaces = false;
      }

      ++str;
   }

   return numWords;
}

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