Question

它忽略了辅音。它忽略了任何类型的空间。它忽略了案例。它唯一不能忽视的是另一个元音是否发生故障。

这些计算：

AEIOU,
aeiou,
hahehihohu,
Take it out

这些不是：

AEIuO,
Taco is good,
Take it over

这是我到目前为止所做的：

import java.util.Scanner;

public class AEIOU_Counter {

public static void main(String[] args) throws Exception {
    java.io.File file = new java.io.File("vowels.txt");

    Scanner input = new Scanner(file);
    String fileContent = "";
    while (input.hasNext())
    {
        fileContent += input.next() + " ";
    }
    input.close();

    char[] charArr = fileContent.toCharArray();
    int counter = 0;
    for (char c : charArr)
    {
        if(c == 'a' || c == 'e' ||c == 'i' ||c == 'o' ||c == 'u')
        counter++;
    }
    System.out.println("The file " + file + " has AEIOU in order " + counter + " times");
}
}

问题是输出：文件vowels.txt按顺序有AEIOU 50次

但是，文件vowels.txt包含：

AEIOU aeiou baeiboeu bbbaaaaaa beaeiou caeuoi ajejijoju aeioo 
aeiOu ma me mi mo mu take it OUT!

所以正确的输出应该是：

The file vowels.txt has AEIOU in order 8 times

Answer 1

这是我能想到的两种方式。没有真正的代码，因为这是你的任务:)

第一种方法是将输入编辑为尽可能简单。

1. Read input from file
2. toLowerCase() the input (to make "aEiOU" simplar as just "aeiou")
3. Remove all non-vowel characters. (so that 'hahehihohu' becomes 'aeiou')
4. Search for literal string "aeiou" and count occurrances.

第二种方法是单独保留输入，但使用循环和计数器。＆＃39;序列＆＃39;可以是数组，也可以是链表

sequence = [a,e,i,o,u] // (or a->e->i->o->u)
curr_char_of_sequence = 'a'
counter = 0

for each char in the input, loop {
    if the char is not a vowel {
        continue to next char
    }

    //see if the vowel is the one we want next
    if char == curr_char_of_sequence {
        //it is! update whats the next vowel we want.
        // ie, if we were looking for an 'a', now look for an 'e'
        curr_char_of_sequence = sequence.next 

        //check to see if we reached the end of the sequence, if so, we found a completed 'aeiou' set
        if curr_char_of_sequence == invalid {
           counter++
           curr_char_of_sequence = 'a'
       }

    //we found a vowel that isn't the right one, restart the sequence
    } else {
        curr_char_of_sequence = 'a'
    }
}

Answer 2

正如人们所指出的，你应该使用正则表达式。这是一个小小的帮助，以这个特定的顺序为你提供每一个AEIOU（不会忽略其间的非元音）

>>> pattern = r'\b.*\b'.join(w)
>>> regex = re.compile(pattern)
>>>
>>> bool(regex.search(s))
True
>>> s = " you who the heck are"
>>> bool(regex.search(s))
False

Answer 3

当然，它每次都是这些字母中的一个。（或者或或......）

尝试使用布尔值，如果你有一个字母（“a”），你会寻找下一个（“e”）。

for (char c : charArr)
{
    if(c == 'a') {
        boolA = true;
    }
    else if(c=='e') {
        if (boolA) {
            boolE = true;
        }
        else {
            boolA = false;
            boolE = false;
            boolI = false;
            boolO = false;
            boolU = false; 
        }
    }
    else if (c=='i') {
        if (boolE) {
            boolI = true;
        }
        else {
            boolA = false;
            boolE = false;
            boolI = false;
            boolO = false;
            boolU = false;  
    }
    //etc, etc ....
}

如果你明白我的意思^^

或者，还有另一种方式（对于懒惰的家伙）你记得你找到的最后一个valide字符，如果真实的字母跟着它，它就赢了。

char lastChar;
String validLetters = "aeiou";
String myArray = "eiou"; //i removed the first

for (char c : charArr) {
    if (c=='a') {
        lastChar=='a';
    }
    else if ( validLetters.contains(c) && lastChar==validLetters.charAt(myArray.indexOf(c)) ) {
        lastChar = c; //u understand it, u get the answer ^^
    }
    else {
        lastChar='w' //just a random char, not in AEIOU
}

最后一个更好，希望它有所帮助，再见：）

Answer 4

看起来我好人打败了我，但这里有一个布尔想法的完整例子

public static void main(String[] args) {

    boolean a = false;
    boolean e = false;
    boolean i = false;
    boolean o = false;
    boolean u = false;

    int vowelCounter = 0;

    String s = "AEIOU aeiou hahehihohu Take it out";

    for (int index = 0; index < s.length(); index++) {
        Character c = Character.toLowerCase(s.charAt(index));
        if (c == 'a') {
            a = true;
            continue;
        }
        if (a && c == 'e') {
            e = true;
            continue;
        }
        if (a && e && c == 'i') {
            i = true;
            continue;
        }
        if (a && e && i && c == 'o') {
            o = true;
            continue;
        }
        if (a && e && i && o && c == 'u') {
            u = true;
            // no continue because we want to exit this if-chain
        }

        if (a && e && i && o && u) {
            vowelCounter++;
            a = e = i = o = u = false; // reset
        }
    }

    System.out.printf("The string \"%s\" contains 'aeiou' in order %d times.\n", s, vowelCounter);
    // The string "AEIOU aeiou hahehihohu Take it out" contains 'aeiou' in order 4 times.
}

读取一个文件并计算元音（字符）＆＃34; aeiou＆＃34;按顺序显示

4 个答案: