不带全零的因子打印数组

Question

i am having an issue trying to get the array that is output from testperfect method to print correctly, i know i need to change my print statement but am unsure how(this statement is in the last method printFactors) i need it to print the factors in the array testperfect makes but i do not want it to print the 0s. I have to use an array and the array has have 100 for its size.

import java.util.Scanner;
public class name_perfect
{
 public static void main ( String args [] )
 {
  int gN;
  int gP = getPerfect();
  int [] array = new int[100];
  boolean tP = testPerfect(gP, array);
  printFactors(gP, array, tP);
  //System.out.println(Arrays.toString(array));
}


public static int getNum() //asks for how many numbers to test
{
 Scanner input = new Scanner ( System.in );
 System.out.print( "How many numbers would you like to test? " );
 int count = input.nextInt();
 int perfect = 1;
 boolean vN = validateNum(count, perfect);
 while(!vN)
 {
  System.out.print (" How many numbers would you like to test? ");
  count = input.nextInt();
  vN = validateNum(count, perfect);
 }
 return count;
 } 

public static boolean validateNum( int count, int perfect  ) //Checks if numbers input are valid
{
 if (( count <= 0) || ( perfect <= 0))

 { 
  System.out.print( "Non-positive numbers are not allowed.\n");
 }



 else 
 {
  return true;
 }
 return false;


}
public static int getPerfect() //asks for the numbers to test
{
 Scanner input = new Scanner ( System.in );
 int perfect = -1;
 int count = getNum();
 System.out.print("Please enter a perfect number: " );
 perfect = input.nextInt(); 
 boolean vN = validateNum(perfect, count);
 while (!vN) 
 {
  System.out.print("Please enter a perfect number: ");
  perfect = input.nextInt();
  vN=validateNum(perfect, count);
 }
 return perfect;
 }


public static boolean testPerfect( int perfect, int[] array ) //tests the numbers that were input 

{
 //testPerfect(perfect, array);
 int limit = perfect;
 int index = 0;
 for ( int i = 1; i < limit ; i++)
 {
  if ( perfect % i == 0)
   { array[i]=i;}
 }

 array[index] = perfect;

 int sum = 0;
 for ( int i = 1; i < limit; i++)
 {
  sum = sum + array[i];
 }

 if ( sum == perfect)
 {
  //Something has to change the array here.
  return true;  
 }

 else
 {
 return false;
 }


}

public static void printFactors(int perfect, int [] array, boolean tP )
 {
 if ( tP == true)
 {
 System.out.println (perfect + ":" + (Arrays.toString(array)));
 }
 else
 {
 System.out.println (perfect + ":" + "NOT PERFECT");
 }

}




}

Answer 1

For this you can have two solutions.

1.Use any Sorting technique or Use the collection framework sorting method and sort this array and then, Iterate through the array, and if the element is '0', dont print that.

    for(int i=0;i<array.length;i++){
       if(array[i]==0)
         continue;
    else
       System.out.println(array[i]);
    }

2.Use an ArrayList. It is flexible to use. Add the perfect numbers to this ArrayList and print that. It will contain only the elements added.

 for(int i=0;arrayList.size();i++)
 {
     System.out.println(arrayList.get(i));
 }

ArrayList is most preferable for this because,

There is no need to initialise how many elements you want. It is flexible to use. It will expand when you add the elements. No need to worry about the size.

Answer 2

I think the below program works fine for your case.

import java.util.ArrayList;
import java.util.Arrays;

public class TestArray {

    private static void print(Integer[] array)
    {
        for(Integer el:array)
            System.out.print(el + " ");
        System.out.println("");
    }


    public static void main(String[] args) {

        // Constructing an Integer array. You can use input-reader to fill the array
        Integer[] array =  new Integer[]{0,1,123,456,-89,0,-1,567,0,23,231,987,0,987654,0};

        // printing the unsorted content 
        print(array);

        // sorting the array
        Arrays.sort(array);

        // printing the sorted content, with '0' values in it 
        print(array);

        //Converts the integer[] to ArrayList, so that i can easily remove elements with value '0' 
        ArrayList<Integer> arrayList = new ArrayList<Integer>(Arrays.asList(array));

        // prints the arraylist
        System.out.println(arrayList);

        //removes the elements with value '0' 
        while(arrayList.contains(new Integer(0)))
            arrayList.remove(new Integer(0));
        System.out.println(arrayList);

        // converting back to Integer[]
        array = arrayList.toArray(new Integer[]{});

        // prints to see that the final array is sorted and conatins no '0' values
        print(array);
    }
}

Please test and let me know the status :)

And the output

0 1 123 456 -89 0 -1 567 0 23 231 987 0 987654 0 
-89 -1 0 0 0 0 0 1 23 123 231 456 567 987 987654 
[-89, -1, 0, 0, 0, 0, 0, 1, 23, 123, 231, 456, 567, 987, 987654]
[-89, -1, 1, 23, 123, 231, 456, 567, 987, 987654]
-89 -1 1 23 123 231 456 567 987 987654 

EDIT

I got your issue solved. Here is another example which just uses the int[] and removes the 0 elements from it.

import java.util.Arrays;

public class TestArray {

    private static void print(int[] array)
    {
        for(int el:array)
            System.out.print(el + " ");
        System.out.println("");
    }


    public static void main(String[] args) {

        // Constructing an Integer array. You can use input-reader to fill the array
        int[] array =  new int[]{0,1,123,456,-89,0,-1,567,0,23,231,987,0,987654,0};
//      int[] array =  new int[]{1,123,456,-89,-1,567,23,231,987,987654};
//      int[] array =  new int[]{1,123,456,-89,-1,567,23,231,987,987654,0};
//      int[] array =  new int[]{1,123,456,567,23,231,987,987654};

        // printing the unsorted content 
        print(array);

        // sorting the array
        Arrays.sort(array);

        // printing the sorted content, with '0' values in it 
        print(array);

        // finds the starting and ending position of '0' elements in the sorted array
        int startPos = -1;
        int endPos = -1;
        for(int i=0;i<array.length;i++)
        {
            if(array[i]==0 && startPos == -1) startPos = i;
            if(array[i]==0) endPos = i;
        }
        System.out.println("startPos : " + startPos + " endPos : " + endPos);

        int[] newArray = null;
        if(startPos!=-1) // there are '0' elements in the array
        {
            // creates another array with new length. 
            int newArrayLength = array.length - (endPos-startPos+1);
            System.out.println("array.length : "+array.length);
            System.out.println("newArrayLength : "+newArrayLength);

            newArray = new int[newArrayLength];
            // copy the contents from original array till start of first '0' value
            System.arraycopy(array, 0, newArray, 0, startPos);
            // copy the remaining contents from original array after end of last '0' value
            System.arraycopy(array, endPos+1, newArray, startPos , newArrayLength-startPos);
        }
        else // no '0' values present in array
        {
            // just copy the original array to new array
            int newArrayLength = array.length ;
            newArray = new int[newArrayLength];
            System.arraycopy(array, 0, newArray, 0 , newArrayLength);
        }

        // prints to see that the final array is sorted and conatins no '0' values
        print(newArray);
    }
}

and the output

0 1 123 456 -89 0 -1 567 0 23 231 987 0 987654 0 
-89 -1 0 0 0 0 0 1 23 123 231 456 567 987 987654 
startPos : 2 endPos : 6
array.length : 15
newArrayLength : 10
-89 -1 1 23 123 231 456 567 987 987654 

Answer 3

  import java.util.*;
  public class TestPerfect {
public static void main(String a[]){
    ArrayList<Integer> perfectNos = new ArrayList<Integer>();
    System.out.println("Enter the perfect number");
    Scanner sc = new Scanner(System.in);
    int per = sc.nextInt();
    int perflag = 0;
    int sum = 0;
    for(int i=1;i<per;i++){
        if(per % i == 0)
            sum =sum+i;
    }
    if(sum == per){
        for(int i=1;i<per;i++)
            if(per % i == 0){
                perfectNos.add(i);
            }
        System.out.print(per+":");
        for(int i=0;i<perfectNos.size();i++)
            System.out.print(perfectNos.get(i)+" ");
    }
    else
        System.out.println(per+": NOT PERFECT");
}

}

This will solve your problem.