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How can I tell my code that it has a "Flush"?

I'm suppose to create a code that recognizes if my hand has the same card faces
public static boolean sameFace(String hand) {
hand = "s9s7s2sQsK";
char f = hand.charAt(0);
if( hand.charAt(0)==hand.charAt(2) && hand.charAt(0)==hand.charAt(4)
&& hand.charAt(0)==hand.charAt(6) && hand.charAt(0)==hand.charAt(8));
return (hand.charAt(0) == hand.charAt(2) && hand.charAt(0) == hand.charAt(4)
&& hand.charAt(0) == hand.charAt(6) && hand.charAt(0) == hand.charAt(8));
sameface = hand;
if (hand==true;)
return (hand==true;) ;
}
As can be seen above, if all positions are the same characters, it comes true(False, if even one isn't the same.) How can I then use the result of that "return" to let my program recognize it has the same faces or not? If that is even possible.
From what i know, based on my code, it's saying "Yes, positions x=y=z are the same" how can I then tell it "Since they are the same, they have the same card faces."
I tried to put this at the end
sameface = hand;
if (hand==true;)
return (hand==true;) ;
Basically I'm trying to say that when the "hand" return statement is true, then samefaces is true. Meaning that the faces are the same. And if it's false it'll return false.

Basically I'm trying to say that when the "hand" return statement is true, then samefaces is true. Meaning that the faces are the same. And if it's false it'll return false.
You do that simply by returning the result of the expression:
public static boolean sameFace(String hand) {
char f = hand.charAt(0);
return f == hand.charAt(2) &&
f == hand.charAt(4) &&
f == hand.charAt(6) &&
f == hand.charAt(8);
}
Or if you want to be friendly to a different number of cards, use a loop:
public static boolean sameFace(String hand) {
char f = hand.charAt(0);
for (int i = 2, len = hand.length(); i < len; i += 2) {
if (f != hand.charAt(i)) {
// Not a match
return false;
}
}
// All matched
return true;
}

How to handle bad input

I'm trying to write a code that determines whether two inputs of DNA sequences are reverse compliments or not. The program asks the user to provide the sequences as a string.
I have the code executing properly but I want to write a single if statement that continues the program if the characters are all 'A' 'T' 'C' or 'G'.
This is what i came up with on my own, but it doesnt work, and it doesn't even look close. I'm new to the language and come from ADA and am just stumped any help would be great.
if ( seqFirst.charAt(i) != 'A' || seqFirst.charAt(i) != 'T' ||
seqFirst.charAt(i) != 'C' || seqFirst.charAt(i) != 'G' ||
seqSecond.charAt(i) != 'A' || seqSecond.charAt(i) != 'T' ||
seqSecond.charAt(i) != 'C' || seqSecond.charAt(i) != 'G' )

You simply need to change || to && throughout the conditional expression.
By way of an explanation, consider this simplified version of your code:
if (c != 'A' || c != 'T' ) { // IS BAD }
and consider the case where c is 'A'. The first predicate evaluates to false. The second predicate evaluates to true. The entire expression is false || true ... which is true ... "BAD"
Now change the || to && and you get false && true ... which is false ... "NOT BAD"
I'm new to the language and come from ADA ...
That's not the real problem. The problem is understanding how boolean algebra works; i.e. DeMorgan's Laws.

As i know, 2 DNA string are reverse compliments when one of these equals to another, but reversed and with changed nucleotides. Since i dont know your version of Java, i wrote some readable method in Java7:
public static boolean isComplimentaryReverse(String str1, String str2) {
//invalid input, could throw exception
if (str1.length() != str2.length()) {
return false;
}
//could be static final field
Map<Character, Character> replaceTable = new HashMap<>();
replaceTable.put('G', 'C');
replaceTable.put('C', 'G');
replaceTable.put('T', 'A');
replaceTable.put('A', 'T');
String reverseStr1 = new StringBuilder(str1).reverse().toString();
for (int i = 0; i < str2.length(); i++) {
//invalid input, could throw exception
if (!replaceTable.containsKey(reverseStr1.charAt(i))) {
return false;
}
if (str2.charAt(i) != replaceTable.get(reverseStr1.charAt(i))) {
return false;
}
}
return true;
}

Simplify with a regular expression.
private static final String VALID_DNA = "[ATCG]+";
...
if (seqFirst.matches(VALID_DNA) && seqSecond.matches(VALID_DNA)) {
// keep going...
}

I'm seeing a possible De Morgan's Law issue. Remember !(a or b) = !a and !b.

how to declare a boolean and only gives true if the first and the last character is "a" [duplicate]

This question already has answers here:
How do I compare strings in Java?
(23 answers)
Closed 7 years ago.
Given a string personName, I'm trying to create a boolean condition2 equal to the condition
the first or last letter in personName is 'A' (case-insensitive). e.g., 'aha' or 'A'
Here's what I've tried so far:
boolean condition2;
if (personName.charAt(0) = "a" || personName.charAt(personName.length()-1) = "a") {
condition2 = true;
} else {
condition2 = false;
}

char type in Java is a character so wouldn't you be looking for something like this?
boolean condition2;
if((personName.charAt(0) == 'a' || personName.charAt(0) == 'A') &&
(personName.charAt(personName.length()-1) == 'a' || personName.charAt(personName.length()-1) == 'A'))
{
condition2 = true;
}
else{
condition2 = false;
}

For your first question where first and last character of a String should be 'a'
boolean condition1 = false;
if(personName.charAt(0) == 'a' && personName.charAt(personName.length()-1) == 'a') {
condition1 = true;
}
For your second condition where variable should be true only if age is in the range [18,24]
boolean condition2 = false;
if(personAge >=18 && personAge <=24) {
condition2 = true;
}

You can do it this way except comparison operator is == and you compare characters, not String.
So right way would be:
Character.toLowerCase(personName.charAt(0)) == 'a'
See, double equals and single quote.

use single quote for data type char. and convert it to lower case, so that it allow whether it's upper case or lower case.
Also use == when comparing if it's equal. this = sign, is for assigning value
if(Character.toLowerCase(personName.charAt(0)) == 'a' ||
Character.toLowerCase(personName.charAt(personName.length()-1)) == 'a')
it will look like this
boolean condition2;
if(Character.toLowerCase(personName.charAt(0)) == 'a' || Character.toLowerCase(personName.charAt(personName.length()-1)) == 'a')
{
condition2 = true;
}
else{
condition2 = false;
}

Your problem can be solve in this manner;
public static void main(String[] args) {
String s = "bsdadasd";
boolean condition2;
System.out.println(check(s.toLowerCase().charAt(0),s.toLowerCase().charAt(s.length()-1)));
}
public static boolean check(char a,char b){
return (a == 'a' || b == 'a');
}
You can pass the two characters as parameters for a method where it return true or falsedepending on the condition.
Since both characters are irrelevant of its case first made them to lowercase. toLowerCase() then passed the char at 0 and char at last.
The return statement will return the true or false to you.
And also use == to check if similar = means assigning.

It is circuitous to write
boolean b;
if (some boolean expression)
b = true;
else
b = false;
Much simpler to write
boolean b = some boolean expression;
For reasons I don't understand, there is a widespread reluctance to write a boolean expression (as distinct from the simple literal values true/false) outside an 'if' statement.
And don't get me started on if (b == true)

How to compare three boolean values

Compare three boolean values and display the first one that is true.
Hey guys, I am trying to write a program that compares three boolean values and displays the first true one. I am comparing three words for their length, and it will display the longest. The error that I am getting is that my else tags aren't working. Take a look at the code.
//Check which word is bigger
if (len1 > len2)
word1bt2 = true;
if (len2 > len3)
word2bt3 = true;
if (len1 > len3)
word1bt3 = true;
//Check which word is the longest
if (word1bt2 == true && word1bt3 == true);
System.out.println(wor1);
else if (word2bt3 == true);
System.out.println(wor2);
else System.out.println(wor3);
I have set boolean values for word1bt2, word2bt3 and word1bt3. In eclipse, I am getting a syntax error under the elses in my code above. Any help would be great!

if (word1bt2 == true && word1bt3 == true);
Is wrong, you need to remove the semicolon:
if (word1bt2 == true && word1bt3 == true)
Same for the elses
else (word2bt3 == true);
Is wrong too, it should be
else if (word2bt3 == true)
Side note: boolean values can be used as condition, so your if statements should be
if (word1bt2 && word1bt3) // The same as if (word1bt2 == true && word1bt3 == true)

How to compare three boolean values?
Dont!
If you find yourself needing to compare three variable you may as well cater for any number of variables immediately - there's no point hanging around - do it properly straight away.
public String longest(Iterator<String> i) {
// Walk the iterator.
String longest = i.hasNext() ? i.next() : null;
while (i.hasNext()) {
String next = i.next();
if (next.length() > longest.length()) {
longest = next;
}
}
return longest;
}
public String longest(Iterable<String> i) {
// Walk the iterator.
return longest(i.iterator());
}
public String longest(String... ss) {
// An array is iterable.
return longest(ss);
}

Remove the ; and change it with brackets {}.
if (word1bt2 && word1bt3) {
System.out.println(wor1);
} else if (word2bt3) {
System.out.println(wor2);
} else {
System.out.println(wor3);
}

Issue with the else blocks: use {} insteaad of () to enclose instructions...
Remove the ; at the first if!!!!! - Quite common mistake, with very puzzling results!
//Check which word is the longest
if (word1bt2 == true && word1bt3 == true) { //leave ; and always add bracket!
System.out.println(wor1);
}
else if(word2bt3 == true)
{
System.out.println(wor2);
}
else {
System.out.println(wor3);
}
if you need a condition in an else branch, you have to use if again - plain else won't have such a feature...
ALWAYS use brackets for bodies of if statements, loops, etc!!!
Be extremely careful NOT to use ; in the lines that don't behave well with it:
if statements
for loops
while() {...} loops' while statement

try this, if lenght are equal then s1 is considered as Bigger. Also i have not added null check
public class Test {
public static void main(String[] args) {
String word1 = "hi";
String word2 = "Hello";
String word3 = "Hell";
String Bigger = null;
if(word1.length() >= word2.length() && word1.length() >= word3.length() ){
Bigger = word1;
}else if(word2.length() >= word1.length() && word2.length() >= word3.length()){
Bigger = word2;
}else if(word3.length() >= word2.length() && word3.length() >= word1.length()){
Bigger = word3;
}
System.out.println(Bigger);
}
}

java - recursive program for judging the validity of expression syntax (beginner)

I'm just learning java, and I have an assignment where I have to write a program that checks the validity of expressions about sets. Valid expressions are capital letters, an expression with a tilde in front, and can be combined using + and x as well as with parentheses. I've written a program that almost works, but I can't figure out how to get the binary operators to work with the parentheses.
It may also be that I have approached the problem in the wrong way (trying to validate from left to right, ignoring everything to the left once it's been validated). I can use any help I can get about writing recursive programs for this sort of problem; that is, if you have any pointers for a better way of approaching the problem, that would be incredibly helpful.
For reference, here is the code that I have:
public static boolean check(String expr) {
char spot;
int close=0;
expr = expr.trim();
//base case
if (expr.length() == 1 && expr.charAt(0)>= 'A' && expr.charAt(0) <= 'Z')
return true;
if (expr.charAt(0) == '~') {
if (expr.charAt(1) == 'x' || expr.charAt(1) == '+' || expr.charAt(1) == ')')
return false;
return check(expr.substring(1));
}
if (expr.indexOf('x') > 0 && expr.indexOf('x') > expr.indexOf(')')) {
int x = expr.indexOf('x');
if (check(expr.substring(0, x)) && check(expr.substring(x)))
return true;
}
if (expr.indexOf('+') > 0 && expr.indexOf('+') > expr.indexOf(')')) {
int plus = expr.indexOf('+');
if (check(expr.substring(0, plus)) && check(expr.substring(plus+1)))
return true;
}
if (expr.charAt(0) == '(') {
close = findEnd(expr.substring(1));
if (close < 0)
return false;
if (check(expr.substring(1,close)) && check(expr.substring(close+1)))
return true;
}
return false;
}

I'm not sure why your code is that complex. Recursion for this is pretty simple overall; here's what I'd do:
public static boolean check(String str) {
if(str.equals("")) return true;
if(str.charAt(0).isAlphaNumeric() || str.charAt(0) == '(' || str.charAt(0) == ')') return check(str.substring(1));
return false;
}
Your edge cases are if the string is empty; if this is the case, then the string is valid. If the character doesn't match what you're looking for, return false. Otherwise, check the next character.