I'm trying to write a change color method in Java that accepts an int parameter and changes that color based on that int. Valid colors will be in the range 1 - 6 for the six colors. You may decide which number (1-6) maps to which color. If the value is not in this range, make the circle red.
I am trying to do this without the use of an array list, but I am unsure how. Any ideas?
I've tried:
if(newColor == 1) {
newColor = "yellow";
}
And I get an error message saying 'incompatible types: java.lang.String cannot be converted to int.'
I've also tried:
if(newColor == 1) {
newColor.equals("yellow");
}
And I get an error message stating 'int cannot be dereferenced'.
1 is an int literal. If the compiler allows you to test newColor ==, then that means newColor must be an int variable. Being an int variable, it is only allowed to hold int values.
"yellow" is a String literal. The compiler will not allow you to assign a String value to an int variable. You may only assign String values to String variables.
You're going to need two variables: One to hold the given int value, and one to hold the String result.
Other languages (e.g., Ruby) might let you do it differently, but if you're going to use Java, then you are going to have to work within the Java rules to solve your problem.
Related
I'm working on the following problem:
Gary is an avid hiker. He tracks his hikes meticulously, paying close attention to small details like topography. During his last hike he took exactly n steps.
For every step he took, he noted if it was an uphill, U, or a downhill, D step. Gary's hikes start and end at sea level and each step up or down represents a 1 unit change in altitude. We define the following terms:
A mountain is a sequence of consecutive steps above sea level, starting with a step up from sea level and ending with a step down to sea level.
A valley is a sequence of consecutive steps below sea level, starting with a step down from sea level and ending with a step up to sea level.
Given Gary's sequence of up and down steps during his last hike, find and print the number of valleys he walked through.
For example, if Gary's path is s = [DDUUUUDD], he first enters a valley 2 units deep. Then he climbs out an up onto a mountain 2 units high. Finally, he returns to sea level and ends his hike.
Function Description
Complete the countingValleys function in the editor below. It must return an integer that denotes the number of valleys Gary traversed.
countingValleys has the following parameter(s):
n: the number of steps Gary takes
s: a string describing his path
Input Format
The first line contains an integer , the number of steps in Gary's hike.
The second line contains a single string , of characters that describe his path.
Output Format
Print a single integer that denotes the number of valleys Gary walked through during his hike.
Sample Input
8
UDDDUDUU
Sample Output
1
Below is my implementation in java. It works for the small test cases but not for the big ones.
static int countingValleys(int n, String s) {
//Use a hashmap to keep track of the number of moves.
HashMap<Character,Integer> map = new HashMap();
boolean sea = true;//check if we are at sea level
//if both D and U have the same total no, we are at sea level.
map.put('D',0);
map.put('U',0);
int valleys = 0;//count num of valleys
for(int i = 0; i < n; i++){
char key = s.charAt(i);
//check if we are at sea level
if(map.get('D') == map.get('U')){//<--PROBLEM
sea = true;
}
else
sea = false;
if(sea == true && key == 'D'){//if we are at sea level and our next key is D, we have a valley
valleys += 1;
}
map.put(key,map.get(key) + 1);//add move and update our hashmap
}
return valleys;
}
The problem seems to be at "if(map.get('D') == map.get('U'))", it seems to be returning false for big numbers, can someone tell me why? It works if I assign each map.get() to a variable and compare the variables instead.
I also wrote the exact same thing in javascript using the "new Object()" type and it passed all the test cases, but it is not working in java with hashmap, why is that?
link to original problem - https://www.hackerrank.com/challenges/counting-valleys/problem?h_l=interview&playlist_slugs%5B%5D=interview-preparation-kit&playlist_slugs%5B%5D=warmup
First, as mentioned in other answer, use .equals() instead of == in this case. An even better approach is, you don't even need to use a Map. Just one integer will be good enough.
As your question is ...returning false for big numbers, can someone tell me why?
Here is the reason.
There are several things you need to understand
1. Types of variable
First, you need to know there are two types of variable in Java: Primitive and Reference.
An integer is usually a Primitive, so the variable itself is the integer value: int a = 1234; : a itself is having value 1234.
To compare primitive variable, you should use ==
For reference type, variable itself is a "pointer". In Java there are Wrapper Classes for primitives. For example, Integer is the wrapper for int. So in Integer a = new Integer(1234);, a is not containing value of 1234. It is a pointer pointing to an Integer object reference. Using == on reference type variables does not compare the content, but only check if the pointer value is the same (i.e. check if they point to same object instance)
2. Autoboxing
Starting from Java 1.5 (iirc), there is a feature called auto-boxing (and unboxing), which ease programmer in converting between primitive types and their corresponding wrapper.
In the past, you need to do something like this:
int a = 1234;
Integer intWrapper = new Integer(a);
int b = intWrapper.intValue();
With autoboxing, you just need to write:
int a = 1234;
Integer intWrapper = a;
int b = intWrapper;
And compiler is going to convert it to:
int a = 1234;
Integer intWrapper = Integer.valueOf(a);
int b = intWrapper.intValue();
So far so good?
Final Answer
So the reason why your code works with small number is:
Integer.valueOf() is caching frequently-used value. From API doc:
public static Integer valueOf(int i)
Returns an Integer instance representing the specified int value. If a new Integer instance is not required, this method should generally be used in preference to the constructor Integer(int), as this method is likely to yield significantly better space and time performance by caching frequently requested values. This method will always cache values in the range -128 to 127, inclusive, and may cache other values outside of this range.
Because it is caching the wrappers, therefore if you are doing map.put(key,map.get(key) + 1), result of get(key) + 1, which is an int, when converting to Integer and if it is a small number, will be a same instance of Integer for same int value. That means, == still works (as variables are pointing to same Integer). However if it is a not-cached number, every invocation will be a different instance, and == will not work (as variables are pointing to different instances of Integer, though the value in the Integer instances are the same)
Suggestion to your algorithm, though a bit off topic:
Your logic is over complicated. It can be greatly simplified to (pseudo code):
countValley(String s) {
currentLevel = 0
valleyCount = 0
for (step in s) {
if step == 'U' {
++currentLevel;
if (currentLevel == 0) { // returning to sea level
++valleyCount
}
} else if step == 'D' {
--currentLevel;
}
}
return valleyCount
}
to compare wrapper Class need to use .equals.
can use Objects.equals :
java.util.Objects.equals(firstInteger, secondInteger);
for case above :
if(java.util.Objects.equals(map.get('D'), map.get('U'))){
sea = true;
}
First, don't use '=='. Use .equals for all classes extends Object.
to compare content of wrapper classes (Integer, Float, Long, Double), use .equals(). "==" would compare the references of both objects. The references might be different even though both contain the same content. You will can however use "==" when you are comparing value types such as int, float, long, double.
Solution 1: If you must use a HashMap, looks like iteration in HashMap is the problem. Below a link to how to do it. As you iterate, build your logic for counting hills and valleys. How to efficiently iterate over each entry in a Java Map?.
Solution 2: Use Array instead of HashMap. After building it, find the D sequences and so on.
i create hash table in java but there is aproblem in it when adding elements in it.
Hashtable <Integer,String> hashT=new Hashtable<Integer ,String >();
hashT.put<sum2 , g>;
the error in second line where define sum2 is variable contain integer value and g variable contain string value. i dont know where the problem in putting elements in hash table.i want to add the values of this variables in hash table each time as the values change.
To mutate an object, you have to call its properties. An object property is either a field or a method. In your case put is a method of the hashT object. A method call is done by writing the object name, then the dot operator, then the method name and finally the arguments surrounded by parenthesis:
objectName.methodName(argument1, argument2, ...);
The problem is here:
hashT.put<sum2 , g>;
put is a method and to call it you have to surround the arguments (sum2 and g) with parenthesis:
hashT.put(sum2, g);
You need to follow the comments and edit your title so its a meaningful title that will help other people understand what your question is about.
Also, when you get an error, copy the big red text that is displayed in the console (color depends on which editor you are using), also known as the error's "Stack Trace" and past it into your question somewhere. This will help us pinpoint what is going on, and the error title itself will likely give away to us exactly what's wrong.
However, without any context on what the error is and what is before or after those two lines of code it is difficult to determine if you have previously defined sum2 or g as a variable that stores values. I am going to assume you haven't assigned at least one of them, likely g as a variable.
For experimental purposes, try replacing those two lines of code with:
Hashtable <Integer,String> hashT=new Hashtable<Integer ,String >();
hashT.put<0 , "g">;
That is putting zero (0) as an explicit integer and g as an explicit string into the hashtable. If you need to put variables into there, then you need to define them before hand, like this:
int sum2 = 3 + 4;
String g = "Some String";
Hashtable <Integer,String> hashT=new Hashtable<Integer ,String >();
hashT.put<sum2 , g>;
Now the integer value 7 is stored as the hash and the string Some String as the mapped value.
I am new to Java so forgive me if this is a silly question.
First I did the following (this is a section from my code):
public static void team()
{
int score = JOptionPane.showInputDialog("Enter the team score...");
calculate(score);
}
But I kept getting an error saying: "Incompatible types string cannot be converted to int".
So I thought I may need to use parsing. I did the following and it worked:
public static void team()
{
int myScore;
String score = JOptionPane.showInputDialog("Enter the team score...");
myScore = Integer.parseInt(score);
calculate(myScore);
}
I would like to know why there is a need for me to parse the variable "score"? In the first piece of code, I declared the variable score as an integer, so would it not store the input which I expect to be an Integer in the variable score which is then passed into the method calculate. I understand that parsing is when you convert a String into an Integer. However in this case, I do not expect the user to input a string so why is there a need for parsing?
The simple reason is: because JOptionPane.showInputDialog() returns a String, regardless of what you actually want the user to input.
There's no mechanism in Java to do automatic conversion based on the type of your destination variable myScore (though some other languages can do this, e.g. implicit conversions in Scala).
You could (in principle) create a custom dialog that returns an int , e.g. by getting the user to choose from a pulldown list of numbers, and then no parsing would be needed. But otherwise, you have to do parsing explicitly.
The JOptionPane.showInputDialog("Enter the team score..."); returns a String which you tried to store in a int. That is why you are getting the "incompatible types".
See Javadocs for more information.
Because a string is stored differently than an integer. Integers are stored in binary, while as a String is a collection of characters.
Example:
The number 14 would be stored as 1110 in Binary. However, as a String, it's stored as 00110001 00110100
Take a look at this http://www.roubaixinteractive.com/PlayGround/Binary_Conversion/The_Characters.asp
Integer.parseInt() basically does the dirty work for you, by looking up the characters in a lookup table, and as long as they correlate with proper integer values, it coverts it over to binary form, to be stored in an integer.
I want to define a grid in which I specify an (x,y) coordinate for each point in the grid. So I want to do something like this:
int [][] pt;
for (x=0; x<numX; x=x+1) {
for (y=0; y<numY; y=y+1) {
pt[x][y] = {xval, yval};
}
}
The reason why is because I am mapping the values of an orderly grid to a disorderly grid. The above code of course causes an exception (unexpected token "{").
What is the best way to do what I'm trying to do? Thanks.
Two things:
You havent initialized your array (maybe you did just didnt put in code)
You are trying to put two values into a place where only one can be held.
Initialize your array like this (if you didnt)
int[][] pt = new int[numX][numY];
To store both values in the array you will need to use an object. The java Point class would be an example of something you could use
Point[][] pt = new Point[numX][numY];
for (x=0; x<numX; x=x+1) {
for (y=0; y<numY; y=y+1) {
pt[x][y] = new Point(xval, yval);;
}
}
You basically want to store a fixed number of values inside every array cell?
Then you are limited with 2 major cases:
Use an object
Java doesn't have user defined value types, so you are forced to use full-blown objects on the heap (with little hope that JVM will be very clever and optimize it, but chances are near zero), be it an array, or any other class.
If both of your values are less than 64 bits, you can pack them in built-in primitive type (such as long) using bitwise arithmetic. (You must be very careful here)
ints are 32 bit, so you can pack 2 ints in 1 long.
pt[x][y] = {xval, yval} is illegal, pt[][] is a double dimensional array. It only can store one value. Just like this pt[x][y] = value
You may try java map.
I am Java beginner and I am trying to code something. Currently I am stucked at getPixelColor(). It's the method of class Robot. I can get pixel color but don't know how to compare with other color. Let's say that my other color is stored in some int variable and I need to compare these two colors. But I can't compare these two colors, because it throws this error "incomparable types:int and java.awt.Color". So how convert it to int?
Thanks
Probably something like:
if(color.getRGB() == stored_color){
}
Anyways, you need to access the Color object's getRGB function to compare the numeric value to another int.
Try
Color otherColor = new Color(someInt);
and then
if (otherColor.equals(robot.getPixelColor(someX, someY))
{
...
}