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Write a program that follows a binary tree in
postorder. Binary Tree will read from a file whose
name will be given as command line parameter. result
browsing will be added at the end of the same file
how can i do this in java
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Member 12069256 wrote: how can i do this in java I fear you will have to work and learn java too.
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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rely my problem just with read tree element from file and write post:order to the same file
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If you follow that link that I gave you you will find samples on all the issues in your question.
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Your homework is set to test what you know, not what a random bunch of people on the Internet know.
Try it yourself, and you'll probably find it's not as difficult as you think.
If you genuinely don't know where to start, then talk to your teacher.
You should also assume that your teacher is monitoring these forums. Even if you managed to persuade someone to do your homework for you, you'd immediately be penalised for cheating.
"These people looked deep within my soul and assigned me a number based on the order in which I joined."
- Homer
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I want to handle 404 page not found exception in my Spring MVC web app, I'm using SPRING 4.2.5.RELEASE, I had read several question regarding this topic but the similar questions are using a different spring java configuration.
I have a Global Exception Handler Controller class that have all my Exceptions, this class works fine but I can't handle a 404 page not found exception.
This is the approach that I take following a tutorial
1) I created a class named ResourceNotFoundException that extends from RuntimeException and I putted this annotation over the class definition
@ResponseStatus(HttpStatus.NOT_FOUND)
like this:
@ResponseStatus(HttpStatus.NOT_FOUND)
public class ResourceNotFoundException extends RuntimeException {
}
2) I created this method in my exception's controller class
@ExceptionHandler(ResourceNotFoundException.class)
@ResponseStatus(HttpStatus.NOT_FOUND)
public String handleResourceNotFoundException() {
return "notFoundJSPPage";
}
But still when I put a URL that doesn't exist I get this error "No mapping found for HTTP request with URI"
The questions that I had read said that I need to enable to true an option for the Dispatcher but since my configuration it's different from the other questions and I don't have a Web.xml I couldn't apply that.
Here it's my Config.java
@EnableWebMvc
@Configuration
@ComponentScan({"config", "controllers"})
public class ConfigMVC extends WebMvcConfigurerAdapter {
@Override
public void addResourceHandlers(ResourceHandlerRegistry registry) {
registry.addResourceHandler("/resources/**").addResourceLocations("/WEB-INF/resources/");
}
@Bean
public UrlBasedViewResolver setupViewResolver() {
UrlBasedViewResolver resolver = new UrlBasedViewResolver();
resolver.setPrefix("/WEB-INF/jsp/");
resolver.setSuffix(".jsp");
resolver.setViewClass(JstlView.class);
return resolver;
}
}
Here is my WebInitializer
public class WebInicializar implements WebApplicationInitializer {
@Override
public void onStartup(ServletContext servletContext) throws ServletException {
AnnotationConfigWebApplicationContext ctx = new AnnotationConfigWebApplicationContext();
ctx.register(ConfigMVC.class);
ctx.setServletContext(servletContext);
Dynamic servlet = servletContext.addServlet("dispatcher", new DispatcherServlet(ctx));
servlet.addMapping("/");
servlet.setLoadOnStartup(1);
}
}
Here is my Global Exception Handler Controller
@ControllerAdvice
public class GlobalExceptionHandlerController {
@ExceptionHandler(value = NullPointerException.class)
public String handleNullPointerException(Exception e) {
System.out.println("A null pointer exception ocurred " + e);
return "nullpointerExceptionPage";
}
@ResponseStatus(value = HttpStatus.INTERNAL_SERVER_ERROR)
@ExceptionHandler(value = Exception.class)
public String handleAllException(Exception e) {
System.out.println("A unknow Exception Ocurred: " + e);
return "unknowExceptionPage";
}
@ExceptionHandler(ResourceNotFoundException.class)
@ResponseStatus(HttpStatus.NOT_FOUND)
public String handleResourceNotFoundException() {
return "notFoundJSPPage";
}
}
And the class I created that extends Runtime Exception
@ResponseStatus(HttpStatus.NOT_FOUND)
public class ResourceNotFoundException extends RuntimeException{
}
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hi everyone
i am new to this site. i am presently doing a practice java program to find reverse numbers, smallest and largest number out of the set of numbers enter and odd and even numbers
my problem is the smallest and largest. the odd and even numbers working okay, the reversed numbers working okay also
i am getting this arrayoutofbound index at line 41 and have been trying to figure out to fix it. can someone help. thanks in advance. here it the codes
another problems also it is repeating and i am not even getting the odd and even numbers to output.
int arr[]= new int[10];
int i=0;
int largest=0, smallest = 0,num;
for(i=0; i<=9; i++)
{
System.out.println("please enter a number");
arr[i] = userinput.nextInt();
}
for(i=9; i>=0; i--)
{
System.out.println(arr[i]+"\t");
}
for(i=0; i<=arr.length; i++)
{
if(i==0)
{
smallest = arr[i];
}
else if(arr[i] < smallest)
{
smallest = arr[i];
}
else if(arr[i]> largest)
{
largest = arr[i];
}
System.out.println("the smallest number is "+smallest);
System.out.println("the largest number is "+ largest);
}
for(i=1; i<=9; i++)
{
if(arr[i]%2==0)
{
System.out.println(arr[i]+" even number");
}
else if(arr[i]%2!=0)
{
System.out.println(arr[i]+" odd number");
}
}
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for(i=0; i<=arr.length; i++)
Your loop should repeat only as long as i < arr.length . The array is 10 items long so index values go from 0 to 9, not 0 to 10.
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thank you very much, I fix but i have one more thing, i need it just to pick out the smallest number and the largest numbers how do i get it to just pick out the smallest number and the largest number instead of this.
the largest number is 205
the smallest number is 0
the largest number is 205
the smallest number is 0
the largest number is 205
the smallest number is 0
the largest number is 205
the smallest number is 0
the largest number is 205
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Put the print statements outside the loop.
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thank you richard got it fix thank much
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A simplified version of your code. And this version don't fail if largest value is the first one.
smallest = arr[0];
largest = arr[0];
for(i=1; i<=arr.length; i++)
{
if(arr[i] < smallest)
{
smallest = arr[i];
}
else if(arr[i]> largest)
{
largest = arr[i];
}
}
System.out.println("the smallest number is "+smallest);
System.out.println("the largest number is "+ largest);
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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thank you so much. got it to work. i never think about that version but thanks all the same.
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Hi everybody,
I am fresh in Java, I would like to be a Java Programmer, so Please advice me what is the required skills, recommended books, latest java technology.
Regards,
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See my answer to the next question below.
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please I want to know more about java. can any one offer an help by telling me what and what I need to know in java. I just concluded my Qbasic programming and next semester we are going into Java. please I need help.
ubonggodwin32@gmail.com or whatapp: 08127725950
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The first, and most important, thing to learn, is how to use Google to help in your research. It would have led you quickly to: The Java Tutorials[^].
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Quote: students have so much free time on their hands that they drew a big jumping game outside the college entrance. There are very large numbers written on the ground as follows:
4 1 5 2 6 3 4 2 0
The number, with a square around it, indicates where you are currently standing. You can jump left or right down the line by jumping the number of spaces indicated by the number you are standing on. So if you are standing on a 4, you can jump either left 4 spaces or right 4 spaces. *BUT: You cannot jump past either end of the line.
For example, the first number (4) only allows you to jump right, since there are no numbers to the left that you can jump to.
The goal: you want to get to the 0 at the far end (right side) of the line. You are also guaranteed that there will be only one zero, which, again, will be at the far right side.
Here is how you do that with the above line:
Starting 4 4 1 5 2 6 3 4 2 0
position
Step 1: 4 4 1 5 2 6 3 4 2 0
Jump right
Step 2: 4 4 1 5 2 6 3 4 2 0
Jump left
Step 3: 4 4 1 5 2 6 3 4 2 0
Jump right
Step 4: 4 4 1 5 2 6 3 4 2 0
Jump right
Step 5: 4 4 1 5 2 6 3 4 2 0
Jump left
Step 6: 4 4 1 5 2 6 3 4 2 0
Jump right
Some FCITgame lines have multiple, correct paths to 0 from the given starting point. Other lines have no paths to 0, such as the following:
1 2 3 0
In this line, you can jump between the 3's, but not anywhere else. So this will return false. You are to write a recursive method that will return a boolean (true or false) to show if we can solve the game (if we can get to the rightmost 0). If we can get to the zero, then your method should return true. If we cannot get the rightmost 0, then the method should return false.
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Very interesting, please let us know if you have a question.
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I could understand this problem any one can help me !!
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I suggest you talk to your teacher and explain what you cannot understand and what extra tuition you need.
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HomeWork !
Do you have a question ?
Where are you stuck ?
Patrice
“Everything should be made as simple as possible, but no simpler.” Albert Einstein
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Hello Sir/Madam,
I have a java code while executing iam getting an error that package is not existed. But I have class files of that package too.
can anyone help me out this problem please.
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Sweacha Nlakshmi wrote: can anyone help me out this problem please. Not without some proper detail of the code and exactly what is happening.
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