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问题:
This Java code is giving me trouble:
String word = <Uses an input>
int y = 3;
char z;
do {
z = word.charAt(y);
if (z!='a' || z!='e' || z!='i' || z!='o' || z!='u')) {
for (int i = 0; i==y; i++) {
wordT = wordT + word.charAt(i);
} break;
}
} while(true);
I want to check if the third letter of word is a non-vowel, and if it is I want it to return the non-vowel and any characters preceding it. If it is a vowel, it checks the next letter in the string, if it's also a vowel then it checks the next one until it finds a non-vowel.
Example:
word = Jaemeas then wordT must = Jaem
Example 2:
word=Jaeoimus then wordT must =Jaeoim
The problem is with my if
statement, I can't figure out how to make it check all the vowels in that one line.
回答1:
Your condition is flawed. Think about the simpler version
z != 'a' || z != 'e'
If z
is 'a'
then the second half will be true since z
is not 'e'
(i.e. the whole condition is true), and if z
is 'e'
then the first half will be true since z
is not 'a'
(again, whole condition true). Of course, if z
is neither 'a'
nor 'e'
then both parts will be true. In other words, your condition will never be false!
You likely want &&
s there instead:
z != 'a' && z != 'e' && ...
Or perhaps:
"aeiou".indexOf(z) < 0
回答2:
Clean method to check for vowels:
public static boolean isVowel(char c) {
return "AEIOUaeiou".indexOf(c) != -1;
}
回答3:
How about an approach using regular expressions? If you use the proper pattern you can get the results from the Matcher object using groups. In the code sample below the call to m.group(1) should return you the string you're looking for as long as there's a pattern match.
String wordT = null;
Pattern patternOne = Pattern.compile("^([\\w]{2}[AEIOUaeiou]*[^AEIOUaeiou]{1}).*");
Matcher m = patternOne.matcher("Jaemeas");
if (m.matches()) {
wordT = m.group(1);
}
Just a little different approach that accomplishes the same goal.
回答4:
For starters, you are checking if the letter is "not a" OR "not e" OR "not i" etc.
Lets say that the letter is i. Then the letter is not a, so that returns "True". Then the entire statement is True because i != a. I think what you are looking for is to AND the statements together, not OR them.
Once you do this, you need to look at how to increment y and check this again. If the first time you get a vowel, you want to see if the next character is a vowel too, or not. This only checks the character at location y=3.
回答5:
Actually there are much more efficient ways to check it but since you've asked what is the problem with yours, I can tell that the problem is you have to change those OR operators with AND operators. With your if statement, it will always be true.
回答6:
String word="Jaemeas";
String wordT="";
int y=3;
char z;
do{
z=word.charAt(y);
if(z!='a'&&z!='e'&&z!='i'&&z!='o'&&z!='u'&&y<word.length()){
for(int i = 0; i<=y;i++){
wordT=wordT+word.charAt(i);
}
break;
}
else{
y++;
}
}while(true);
here is my answer.
回答7:
I have declared a char[] constant for the VOWELS, then implemented a method that checks whether a char is a vowel or not (returning a boolean value). In my main method, I am declaring a string and converting it to an array of chars, so that I can pass the index of the char array as the parameter of my isVowel method:
public class FindVowelsInString {
static final char[] VOWELS = {'a', 'e', 'i', 'o', 'u'};
public static void main(String[] args) {
String str = "hello";
char[] array = str.toCharArray();
//Check with a consonant
boolean vowelChecker = FindVowelsInString.isVowel(array[0]);
System.out.println("Is this a character a vowel?" + vowelChecker);
//Check with a vowel
boolean vowelChecker2 = FindVowelsInString.isVowel(array[1]);
System.out.println("Is this a character a vowel?" + vowelChecker2);
}
private static boolean isVowel(char vowel) {
boolean isVowel = false;
for (int i = 0; i < FindVowelsInString.getVowel().length; i++) {
if (FindVowelsInString.getVowel()[i] == vowel) {
isVowel = true;
}
}
return isVowel;
}
public static char[] getVowel() {
return FindVowelsInString.VOWELS;
}
}