标题 “hello everyone nihao!”判断该字符串有多少个单词
#include int main(int argc, char *argv[]){ char *str = "hello everyone nihao!"; int cnt = 0; int word_len = 0; while('\0' != *str) { //用空格区分单词 没有考虑特殊字符 if (' ' == *str) { //避免多个空格连在一起 if (word_len > 0) { cnt++; word_len = 0; } } //单词长度 避免多个空格 如果一个字符是不是不应该算单词？ else { word_len++; } str++; } //最后一个单词之后大多是结束而没有空格 只能通过单词长度来区分是否有单词 if (word_len > 0) { cnt++; } printf("单词个数:%d
", cnt); return 0;}一个比较通用的单词个数统计，但没有考虑特殊字符，以空格区分单词
直接找不含exp字符的单词正则是实现不了的.
不过可以取巧,找到含exp字符单词,然后把这些单词替换成空字符串,然后再在替换好的字符串里面找单词
找含exp的单词的正则
(?=[A-Za-z]*exp[A-Za-z]*)[A-Za-z]*
匹配替换后的字符串的字符的正则
[A-Za-z]*
如图
#include <stdio.h>//c语言
#include <iostream>//c++
int main(){
char[] str = "hello everyone nihao!";
int count = 0;//用于统计空格个数以此来计算单词数量
if(str[0] == 0){//如果字符串为空
fprintf("单词个数为%d",count);//c语言
std::cout<<"单词个数为"<<count<<endl;//c++
return 0;
}
count++;//若字符串不为空则默认至少有一个单词
for(int i=0;;i++){
if(str[i] == 0)//字符串结束标志的ASCII码的值为0
break;
if(str[i] == ' ')//若遇到空格字符
count++;
}
fprintf("单词个数为%d",count);//c语言
std::cout<<"单词个数为"<<count<<endl;//c++
return 0;
}
如果是java的话
class Solution {
public static void main() {
String str = "hello everyone nihao!";
if (str.isEmpty()) {// 若字符串为空
System.out.println("单词个数为0");
return;
}
System.out.println("单词个数是:" + str.split(" ").length);
}
}
如果是C#的话跟java差不多
using System;
class Solution {
public static void main(string[] args) {
string str = "hello everyone nihao!";
if (str.length()==0) {// 若字符串为空
Console.WriteLine("单词个数为0");
return;
}
Console.WriteLine("单词个数是:" + str.split(' ').length);
}
}
Python
if __name__ == "__main__":
str = "hello everyone nihao!"
if str=="":
print("单词个数为0")
print(len(str.split(' ')))
hello.hell.he.every.ever.very.one.on.everyone.9个单词 nihao是拼音，不算单词。
两个啊。“hello”和“everyone”。😂
用sizeof（）运算符不过得减去3，因为有一个是空字符，和两个空格。或者用strlen()函数，减去2，因为有2个空格。
满意请采纳！谢谢
你要判断这个必须建立一个英语词典数据库。根据单词使用量和使用范围进行比对。没有简洁方法。除非你和有翻译软件的翻译公司有一腿，让他们给你开个服务器后门。
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