We Keep Coding

Method to see if a character is in an array of Characters

I have these two methods- My code doesn't seem to be working as planned
What is is supposed to do is- Go through the array of characters- If there's not another character the same in the array, it is supposed to add itself to the index variable-
This is my comparing method
private boolean isValid(char c) {
for(int i = 0; i < letters.length; i++) {
if(Arrays.asList(letters).equals(c)) {
return false; //Not valid
}
}
return true;
Full code is below though
public void generate(String first, String second) {
tempString = new StringBuilder(first+second).reverse();
letters = new char[tempString.length()];
for(int i = 0; i < tempString.length(); i++) {
letters[i]= tempString.charAt(i);
if(isValid(tempString.charAt(i))) {
index += i;
}
}
}
private boolean isValid(char c) {
for(int i = 0; i < letters.length; i++) {
if(Arrays.asList(letters).equals(c)) {
return false; //Not valid
}
}
return true;
}

There is no need to convert to a List (and a List is not a primitive char), you can use == for comparing primitive values (such as your chars). Something like,
private boolean isValid(char c) {
for (int i = 0; i < letters.length; i++) {
if (letters[i] == c) {
return false;
}
}
return true;
}
or with an enhanced for-each loop like
private boolean isValid(char c) {
for (char letter : letters) { // <-- for each letter in letters.
if (letter == c) { // <-- if the letter is equal to the argument.
return false;
}
}
return true;
}
You should also test for validity before adding to the array like
if (isValid(tempString.charAt(i))) {
letters[index] = tempString.charAt(i);
index++;
}

Arrays.asList(letters).equals(c) is comparing the list to the character (Is the list equal to this character which is not what you want.
To find if a character is in the string you can instead do
string.indexOf('a') which will return -1 if the character is not present and >= 0 if it is in the string.

Verify if String matches real number

I'm trying to verify if a String s match/is a real number. For that I created this method:
public static boolean Real(String s, int i) {
boolean resp = false;
//
if ( i == s.length() ) {
resp = true;
} else if ( s.charAt(i) >= '0' && s.charAt(i) <= '9' ) {
resp = Real(s, i + 1);
} else {
resp = false;
}
return resp;
}
public static boolean isReal(String s) {
return Real(s, 0);
}
But obviously it works only for round numbers. Can anybody give me a tip on how to do this?
P.S: I can only use s.charAt(int) e length() Java functions.

You could try doing something like this. Added recursive solution as well.
public static void main(String[] args) {
System.out.println(isReal("123.12"));
}
public static boolean isReal(String string) {
boolean delimiterMatched = false;
char delimiter = '.';
for (int i = 0; i < string.length(); i++) {
char c = string.charAt(i);
if (!(c >= '0' && c <= '9' || c == delimiter)) {
// contains not number
return false;
}
if (c == delimiter) {
// delimiter matched twice
if (delimiterMatched) {
return false;
}
delimiterMatched = true;
}
}
// if matched delimiter once return true
return delimiterMatched;
}
Recursive solution
public static boolean isRealRecursive(String string) {
return isRealRecursive(string, 0, false);
}
private static boolean isRealRecursive(String string, int position, boolean delimiterMatched) {
char delimiter = '.';
if (position == string.length()) {
return delimiterMatched;
}
char c = string.charAt(position);
if (!(c >= '0' && c <= '9' || c == delimiter)) {
// contains not number
return false;
}
if (c == delimiter) {
// delimiter matched twice
if (delimiterMatched) {
return false;
}
delimiterMatched = true;
}
return isRealRecursive(string, position+1, delimiterMatched);
}

You need to use Regex. The regex to verify that whether a string holds a float number is:
^[-+]?[0-9]*\.?[0-9]+([eE][-+]?[0-9]+)?$

Can anybody give me a tip on how to do this?
Starting with your existing recursive matcher for whole numbers, modify it and use it in another method to match the whole numbers in:
["+"|"-"]<whole-number>["."[<whole-number>]]
Hint: you will most likely need to change the existing method to return the index of that last character matched rather than just true / false. Think of the best way to encode "no match" in an integer result.

public static boolean isReal(String str) {
boolean real = true;
boolean sawDot = false;
char c;
for(int i = str.length() - 1; 0 <= i && real; i --) {
c = str.charAt(i);
if('-' == c || '+' == c) {
if(0 != i) {
real = false;
}
} else if('.' == c) {
if(!sawDot)
sawDot = true;
else
real = false;
} else {
if('0' > c || '9' < c)
real = false;
}
}
return real;
}

Palindrome tester

So I have the majority of the code written and it works. Except for the iterative method keeps showing that it is not a palindrome regardless of what is typed in. I am at a loss as to how to remedy it here is the code.
//David Crouse Assignment 2
import java.util.Scanner;
public class Assignment2 {
public static boolean loop = false;
//main
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.println("Welcome to the Palindrome Checker!");
do{
System.out.print("Enter a string to check if it is a palindrome. ");
System.out.print("Enter x to exit.");
String word = input.nextLine();
word = word.replaceAll("\\s","");
word = word.toLowerCase();
//Exit Program
if(word.equalsIgnoreCase("x")){
System.out.println("End of program. Good Bye!");
System.exit(0);
}
if(iterativePalindromeChecker(word)){
System.out.println("Iterative Result: Palindrome!");
}else{
System.out.println("Iterative Result: Not a palindrome");
}
if(recursivePalindromeChecker(word)){
System.out.println("Recursive Result: Palindrome!\n");
}else{
System.out.println("Recursive Result: Not a palindrome\n");
}
loop = true;
}while (loop == true);
}
//Iterative Method
public static boolean iterativePalindromeChecker(String str){
boolean result = false;
int length = str.length();
int i, begin, end, middle;
begin = 0;
end = length - 1;
middle = (begin + end)/2;
for (i = begin; i <= middle; i++) {
if (str.charAt(begin) == str.charAt(end)) {
begin++;
end--;
}
else {
break;
}
}
if (i == middle + 1) {
result = false;
}
return result;
}
//Recusive Methods
public static boolean recursivePalindromeChecker(String str){
if(str.length() == 0 || str.length() == 1)
return true;
if(str.charAt(0) == str.charAt(str.length()-1))
return recursivePalindromeChecker(str.substring(1,str.length()-1));
return false;
}
}

Your iterative method never sets result to be true. Here's a modified version:
public static boolean iterativePalindromeChecker(String str){
int length = str.length();
int i, begin, end, middle;
begin = 0;
end = length - 1;
middle = (begin + end)/2;
for (i = begin; i <= middle; i++) {
if (str.charAt(begin) == str.charAt(end)) {
begin++;
end--;
}
else {
return false;
}
}
return true;
}

Your iterative method does not set result = true anywhere, so it really can't help it. Although I think the iterative method could be better overall. Take a close look at what is happening in the recursive one and see if you can implement some of it (like the guard conditions) more closely in the iterative method, and keep in mind that you are not limited to a single index value in a for loop either. e.g.:
public static boolean iterativePalindromeChecker(String str) {
for(int start = 0, end = str.length() - 1; start < end; start++, end--) {
if(str.charAt(start) != str.charAt(end)) {
return false;
}
}
return true;
}

I'm guessing someone once told you that a function should only have one return point, and trying to follow that led you to using a mutable result variable which screwed you here. Using break poses the same ostensible problem anyway. Save yourself the headache and just return as soon as you know the answer.
public static boolean iterativePalindromeChecker(String str) {
int begin = 0;
int end = str.length() - 1;
while (begin < end) {
if (str.charAt(begin) != str.charAt(end)) {
return false;
}
begin++;
end--;
}
return true;
}

Recognize String has digits

This is my input verifier:
public class MyInputVerifier extends InputVerifier {
#Override
public boolean verify(JComponent input) {
String text = ((JTextField) input).getText().trim()
if (text.isEmpty() || text.length() == 0) return false;
// How verifier that if text contains digit, return false?
return true;
}
I need to recognize when text contains digit(s), it return false to me.
Is there any method Or i should use old way?(for loop)

This is one way of checking if text contains a digit:
boolean containsADigit = text.matches(".*\\d.*");
If you want to check if all characters are digits you can still use regular expressions, or try parsing the string as an integer:
boolean isDigitsOnly = text.matches("\\d*");

Using only very basic stuff:
public boolean containsDigit(String str) {
int n = str.size();
for(int i=1, i<n, ++i) {
if(str.charAt(i).isDigit()) {return true;}
}
return false;
}

Are you looking for a more faster way?
private boolean hasOnlyDigits(String str) {
for(int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if (c < '0' || c > '9') {
return false;
}
}
return true;
}

Translating a regex into a more basic code

How can I write the below code without using regex?
public static boolean validateCode(String code){
boolean hasAtLeastOneNumber = Pattern.compile("[0-9].*[0-9]")
.matcher(code).find();
boolean hasAtLeastTwoLetters = Pattern.compile("[a-zA-Z].*[a-zA-Z]")
.matcher(code).find();
boolean hasAtLeastOneHyphen = Pattern.compile("-")
.matcher(code).find();
}

How about
public static boolean validateCode2(String code) {
int numbers = 0, letters = 0, hyphens = 0;
for (char c : code.toCharArray()) {
if (Character.isDigit(c)) numbers++;
if (Character.isAlphabetic(c)) letters++;
if (c=='-') hyphens++;
}
return numbers>=2 && letters>=2 && hyphens>=1;
}

For hasAtLeastOneNumber:
for (char c : code.toCharArray()) {
if (Character.isDigit(c)) {
return true;
}
return false;
For hasAtLeastTwoLetters:
int numFound = 0;
for (char c : code.toCharArray()) {
if (Character.isLetter(c)) {
numFound++;
if (numFound >= 2) {
return true;
}
}
}
return false;
For hasAtLeastOneHyphen:
for (char c : code.toCharArray()) {
if (c == '-') {
return true;
}
}
return false;
If you don't want to use toCharArray, you could use:
for (int i=0; i<code.length(); i++) {
char c = code.charAt(i);
// do the rest of the test here
}
That's basically equivalent to using toCharArray except that it's slightly more confusing: someone who looks at the code would need to take a second or two to figure it out. With toCharArray it's obvious what you're doing.

You can loop through the string and test it for ranges of characters. See an example on IDEONE, or ask me if you need an explanation.
import java.util.*;
import java.lang.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
System.out.println(validarCodigo("No-numbers"));
System.out.println(validarCodigo("1-A"));
System.out.println(validarCodigo("This 1 Matches -- :-)"));
}
public static boolean validarCodigo(String codigo) {
int i;
char[] chars = codigo.toCharArray();
char current;
boolean tieneAlmenosUnNumero = false;
boolean tieneAlmenosDosLetras = false;
boolean tieneAlmenosUnGuion = false;
// Check for at least one number
for (i=0; i<chars.length; i++) {
current = chars[i];
if (current >= '0' && current <= '9') {
tieneAlmenosUnNumero = true;
break;
}
}
// Check for at least two letters
int found = 0;
for (i=0; i<chars.length; i++) {
current = chars[i];
boolean lower = current >= 'a' && current <= 'z';
boolean upper = current >= 'A' && current <= 'Z';
if (lower || upper) found++;
if (found == 2){
tieneAlmenosDosLetras = true;
break;
}
}
// Check for at least one hyphen
for (i=0; i<chars.length; i++) {
current = chars[i];
if (current == '-') {
tieneAlmenosUnGuion = true;
break;
}
}
return tieneAlmenosUnNumero && tieneAlmenosDosLetras && tieneAlmenosUnGuion;
}
}