public static void main(String[] args){
    int maxNumber = 0;
    int secondMaxNumber = 0;
    if(args.length == 0){
      System.err.println("Number array is empty");
      return;
    }
    for(int i=0; i < args.length; i++){
      int currNumber = Integer.parseInt(args[i]);
      if(maxNumber < currNumber){
        secondMaxNumber = maxNumber;
        maxNumber = currNumber;
      }else if(secondMaxNumber < currNumber){
        secondMaxNumber = currNumber;
      }
    }
    System.err.println("Max. number is "+maxNumber);
    System.err.println("Second Max. is "+secondMaxNumber);
  }
}

public class SecondMaximumNumber{
  public static void main(String[] args){
    int maxNumber = 0;
    int secondMaxNumber = 0;
    if(args.length == 0){
      System.err.println("Number array is empty");
      return;
    }
    for(int i=0; i < args.length; i++){
      int currNumber = Integer.parseInt(args[i]);
      if(maxNumber < currNumber){
        secondMaxNumber = maxNumber;
        maxNumber = currNumber;
      }else if(secondMaxNumber < currNumber){
        secondMaxNumber = currNumber;
      }
    }
    System.err.println("Max. number is "+maxNumber);
    System.err.println("Second Max. number is
"+secondMaxNumber);
  }
}

import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.Set;


public class SecondMax {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		int[] numbers = {9,4,8,0,0,5,9,1,4,2};
		Set arrSet = new HashSet();
		for (int i=0;i<numbers.length;i++) {
			arrSet.add(numbers[i]);
		}
		ArrayList s = new ArrayList(arrSet);
		Collections.sort(s);

		System.out.println("Second element : " + 
s.get(s.size()-2));
	}
}

Himesh Mistry -best solution

Ranjaram is correct.I modified and testes it.
public class test123 {
	public static void main(String[] args){
	    int maxNumber = 0;
	    int secondMaxNumber = 0;
	  
	    int[] anArray;           
	    anArray =new int [10];
	    anArray[0] = 100; 
        anArray[1] = 200; 
        anArray[2] = 300; 
        anArray[3] = 400;
        anArray[4] = 500;

	    if(anArray.length == 0){
	      System.err.println("Number array is empty");
	      return;
	    }
	   for(int i=0; i < anArray.length; i++){
	     int currNumber = anArray[i];
	      if(maxNumber < currNumber){
	        secondMaxNumber = maxNumber;
	        maxNumber = currNumber;
	      }else if(secondMaxNumber < currNumber){
	        secondMaxNumber = currNumber;
	      }
	    }
	    System.err.println("Max. number is "+maxNumber);
	    System.err.println("Second Max. 
is "+secondMaxNumber);
	  }
	}

I think, the except for Himesh's program, all other ones 
doesnt handle duplicate values ...
I mean if the numbers are :- 9 9 8 8 7 7 7 1 3 2
the answer should be 8 ideally.
But I doubt whether thats really happening in any other 
programs, except for Himesh's.
Do let me know, if I am wrong. Thanks.

hey using collections solving this type of problem becomes
easy but whenever u go for interview most of the time they
ask you to write program without using collection API thts
why i think rajaram's code is correct

If we give negative values, what would be the Rajaraman 
code begaves????

This is I tried.
public class SecondLargestNumber {
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		int[] num = new int[]{-101,-105,-2,-7,-22,-
104,-8,-10,-100,-102,-102};
		int big=-1;
		int secbig=-1;
		if(num.length == 1) {
			big = num[0];
			secbig = num[0];
		} else {
			if(num[0] > num[1]) {
				big = num[0];
				secbig = num[1];
			} else {
				big = num[1];
				secbig = num[0];
			}
		} 
		if(num.length > 2){
			for(int i=2;i<num.length;++i){
				if(num[i] > secbig && num
[i] <big) {
					secbig = num[i];
				}
				if(num[i]>=big){
					secbig = big;
					big = num[i];
				}
			}
		}
		System.out.println(big);
		System.out.println(secbig);
	}
}

Please check is it working and let me know. For removing 
duplicates, we will have one more method to remove the 
duplicate elements.