java java元音计数器使用开关方法
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java Vowel Counter Using Switch Method
提问by Mast3rCh1ef
I am supposed to write a simple java application that counts each vowel in a string entered by the user and outputs the number of time each vowel occurs.
我应该编写一个简单的 java 应用程序，它计算用户输入的字符串中的每个元音并输出每个元音出现的次数。
I don't understand why my code is checking each individual word in my string. I am getting the right amount of vowels for each word. Here is what I have:
我不明白为什么我的代码要检查字符串中的每个单词。我为每个单词获得了适量的元音。这是我所拥有的：
import java.util.Scanner;
public class VowelAnalyst
{
//
//
public static void main (String[] args)
{
String userString;
int aCount = 0, eCount = 0, iCount = 0, oCount = 0, uCount = 0;
char vowels;
Scanner scan = new Scanner (System.in);
System.out.println ("enter string:");
userString = scan.nextLine();
for (int count = 0; count < userString.length(); count++)
{
vowels = userString.charAt(count);
switch (vowels)
{
case 'a':
aCount++;
break;
case 'e':
eCount++;
break;
case 'i':
iCount++;
break;
case 'o':
oCount++;
break;
case 'u':
uCount++;
break;
default:
System.out.println ("Please enter valid string.");
}
System.out.println ("a: " +aCount);
System.out.println ("e: " +eCount);
System.out.println ("i: " +iCount);
System.out.println ("o: " +oCount);
System.out.println ("u: " +uCount);
}
}
}
回答by Rohit Jain
May be you should move your below print statements out of your for loop, else they will print count after every character compared: -
可能您应该将下面的打印语句移出 for 循环，否则它们将在比较每个字符后打印计数：-
System.out.println ("a: " +aCount);
System.out.println ("e: " +eCount);
System.out.println ("i: " +iCount);
System.out.println ("o: " +oCount);
System.out.println ("u: " +uCount);
UPDATE: -
更新：-
Although the way you are doing is not a bad way, but you would be better if you maintain a Map<Character, Integer>to store the count of each Vowel. You need to initialize your Map with an initial count of 0 for each character, and then on each character read, just increment the count, if match is found in Map.
虽然你这样做的方式还不错，但是如果你维护一个Map<Character, Integer>来存储每个Vowel. 您需要使用每个字符的初始计数 0 来初始化您的 Map，然后在读取每个字符时，如果在 Map 中找到匹配，只需增加计数。
Here's a sample snippet: -
这是一个示例片段：-
// This is `double-braces` initialization.
// You can rather initialize your Map in a way you are comfortable with
Map<Character, Integer> vowels = new HashMap<Character, Integer>() {
{
put('a', 0);
put('e', 0);
put('i', 0);
put('o', 0);
put('u', 0);
}
}; // Note the semi-colon here
And then your code of reading each character from string in for loop: -
然后你在 for 循环中从字符串中读取每个字符的代码： -
for (int count = 0; count < userString.length(); count++)
{
char ch = userString.charAt(count);
ch = Character.toLowerCase(ch);
if (vowels.containsKey(ch)) {
vowels.put(ch, vowels.get(ch) + 1);
}
}
System.out.println(vowels); // Will print each vowels with respective count
回答by Adrian Adamczyk
Do you guys really have to create a map to store some characters?
你们真的需要创建一个地图来存储一些字符吗？
public static void main(String[] args)
{
char[] vowels = "aeiouAEIOU".toCharArray();
String text = "I am a Java programmer";
int[] vowelsCount = new int[vowels.length];
for (char textChar: text.toCharArray())
{
for (int i = 0; i < vowels.length; i++)
{
char vowel = vowels[i];
if (vowel == textChar)
{
vowelsCount[i]++;
break;
}
}
}
for (int i = 0; i < vowelsCount.length; i++)
{
System.out.println(vowels[i] + " - " + vowelsCount[i]);
}
}
Output:
输出：
a - 5
e - 1
i - 0
o - 1
u - 0
A - 0
E - 0
I - 1
O - 0
U - 0
