验证用户输入仅是斐波那契数列上的数字

Question

import java.util.Scanner;

public class Fibonacci {
public static void main(String args[]) {


    
    int result = UserInput();

    fibArray[0] = 1;
    fibArray[1] = 1;

    for (int i = 0; i < result; i++)


        System.out.print(fibonacci(i) + " ");

}


public static long fibArray[] = new long[1000];

public static long fibonacci(long n) {
    long fibValue = 0;
    if (n == 0) {
        return 0;
    } else if (n == 1) {
        return 1;

    } else if (fibArray[(int) n] != 0) {
        return fibArray[(int) n];
    } else {
        fibValue = fibonacci(n - 1) + fibonacci(n - 2);
        fibArray[(int) n] = fibValue;
        return fibValue;
    }
}

public static int UserInput() {

    System.out.print("Enter a Fibonacci Number ");
    Scanner test = new Scanner(System.in);
    int n = test.nextInt();
   

    return n;
}
}

Hi This is not homework nor is it a college assignment. I'm just learning java and I would like to implement a check in my UserInput method on Fibonacci sequence, I want only numbers to be entered and not special characters or letters. I've read other similar topics here but I'm still a bit unsure as to how to do this.

How would I achieve this?

And btw sorry if the code is sloppy. Thanks

Answer 1

Check this out .

You can modify your code like this. The idea is to check whether the String you are getting from the Scanner is an integer or not (using the Integer#parseInt method). Also the general convention in java is to use lower camel case for the method name.

public static void userInput(){
    System.out.print("Enter a Fibonacci Number");
    Scanner test = new Scanner(System.in);
    String possibleInteger = test.next();
    if(isInteger(possibleInteger)){
      //Continue with your code
    }
}

private static boolean isInteger(String possibleInteger){
    try{
        Integer.parseInt(possibleInteger);
        return true;
    } catch( NumberFormatException ignore ){
        return false;
    }
}

Answer 2

btw, i would approach this with building the fibonacci sequence in an iterative way instead of recursive and memoization (and just check if the current number is bigger than the checked one), that way you don't need the extra space and you can support larger than fib_1000, good luck!

Answer 3

To check if an entered value is part of a fibonacci series

• first add a set to add the fibValue in your method where you add it to the long array

public static Set<Long> fibSet = new HashSet<>();

• then do the following:

   int value = UserInput(); boolean has = fibSet.contains(value); for (int i = 0; i < 1000 && has == false; i++) { if (value == fibonacci(i)) { has = true; break; } } System.out.println(value + " is" + (has? "": " not") + " a term in the standard fibonacci sequence");

  }

Note that the static set is only useful for multiple runs in the same program as is checks for previously added entries.

And the 1000 value in the loop is arbitrary and can be altered to either increase or decrease the number of terms.

Answer 4

This is the solution I have it works with most numbers but can fail if it receives numbers bigger than the number limit for integers. public static boolean checkIfNumber() {

   int pin = 21211;
   Integer pinT = pin;
   int total = 0;
  //Make a character array with all the characters that you want to have.
  char[] numbers = {'1', '2', '3', '4', '5', '6', '7', '8', '9', '0'};
  
  for (int i = 0; i < number.length; i++) {
       if (pinT.toString().contains(numbers[i]) {
       total += 1;
       }
      }
  return total < pintT.toString().length();