Simple Logic of lowest and second lowest number in JAVA.

Display lowest and second lowest number & handle the different exceptions possible to be thrown during execution in JAVA.
LOGIC 1: Use Sort Method
CODING:

import java.util.*;
class lowest
{
 public static void main(String[] args)
 {
  Scanner num=new Scanner(System.in);
   try
    {
     System.out.print("How Many Numbers want to Input:");
     int[] arry=new int[num.nextInt()];
     for(int i=0;i<arry.length;i++)
      {
       System.out.print("Input A Number:");
       arry[i]=num.nextInt();
      }
     Arrays.sort(arry);
     System.out.println("Lowest Number : " + arry[0] + "\nSecond Lowest Number : " + arry[1]);
    } 
   catch(InputMismatchException e1)
    {
     System.out.print("Error::Input only Numeric Value");
    }
  }
}

LOGIC 2:Without use Sort Method,time complexity of logic 2 less than logic 1,because each data is sorting in logic 1
CODING:

import javax.swing.JOptionPane;  //To Use Input Dialog Box
import java.util.*;  //To use NumberFormatException Class
class lowest
{
 public static void main(String[] args)
 {
  String num=JOptionPane.showInputDialog(null, "Enter Number...");
  int lowest=Integer.parseInt(num),Slowest=Integer.parseInt(num);
   try
    {
     for(int i=0;i<n;i++)//'n' means how many numbers user input
      {
       num=JOptionPane.showInputDialog(null, "Enter Number..."); 
        if(Integer.parseInt(num)!=lowest && Integer.parseInt(num)!=Slowest) 
           {
             if(lowest>Integer.parseInt(num))
              {
               Slowest=lowest;
               lowest=Integer.parseInt(num);
              }
             else
              {
                if(Slowest>Integer.parseInt(num) || Slowest==lowest)
                 Slowest=Integer.parseInt(num);
              }
           }
       }
    } 
   catch(NumberFormatException e)
    {
     System.out.print("Error::Input Numeric Value...");
    }
  System.out.println("Lowest Number : " + lowest + "\nSecond Lowest Number : " + Slowest);
  }
}