计算字符串C#数组中每一行的元音

Question

我正在写一个代码，计算字符串数组中最频繁的单词，我还试图计算单词每行的元音数量，换句话说，数组中所有单词的元音总数，我已经写了代码，但它不起作用，我不知道它有什么问题，是因为我在比较字符串和字符吗？

using System;
using System.Collections.Generic;
namespace lab1._2
{
class Program
{
static String findWord(String[] arr)
{
Dictionary<String, int> hs =
new Dictionary<String, int>();
for (int i = 0; i < arr.Length; i++)
{
if (hs.ContainsKey(arr[i]))
{
hs[arr[i]] = hs[arr[i]] + 1;
}
else
{
hs.Add(arr[i], 1);
}
}
String key = "";
int value = 0;
foreach (KeyValuePair<String, int> me in hs)
{
if (me.Value > value)
{
value = me.Value;
key = me.Key;
}
}
return key;
}
static void Main(string[] args)
{
int s;
//char[] v = new char[10] {'a','A','e','E','i','I','o','O','u','U'};
Console.WriteLine("Enter size of array : ");
s = Convert.ToInt32(Console.ReadLine());
string[] arr = new string[s];
Console.WriteLine("Enter string elements : ");
for (int i = 0; i < s; i++)
{
arr[i] = Console.ReadLine();
}
Console.WriteLine("nArray elements : ");
for (int i = 0; i < s; i++)
{
Console.WriteLine(arr[i]);
}
Console.WriteLine("nThe most frequent word : ");
Console.WriteLine(findWord(arr));
int vowel = 0, cons = 0;
for (int i = 0; i < arr.Length; i++)
{
if (arr[i] == "a" || arr[i] == "e" || arr[i] == "i" || arr[i] == "o" || arr[i] 
== "u")
{
vowel++;
}
else
cons++;
}
Console.WriteLine("Vowels : ");
Console.WriteLine(vowel);
Console.WriteLine("Consants : ");
Console.WriteLine(cons);
}
}
}

Answer 1

我认为问题在于arr是一个字符串数组，并且您正在迭代它，就好像它是一个单独的字符串一样。

一个简单的方法是使用嵌套循环。

foreach (var s in arr) // s is a string
{
for (int i = 0; i < s.Length; i++)
{
if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u')
{
vowel++;
}
else
cons++;
}
}

https://dotnetfiddle.net/tI3oTc

using System;
public class Program
{
public static void Main()
{
int vowel = 0;
int cons = 0;
string[] arr = new string[]{"test"};
foreach (var s in arr) // s is a string
{
for (int i = 0; i < s.Length; i++)
{
if (s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u')
{
vowel++;
}
else
cons++;
}
}
Console.WriteLine(vowel);
}
}