return a new array length 2 challenge

[sai rama krishna]

Hi,

I am working on below challenge

Given 2 int arrays, a and b, return a new array length 2 containing, as much as will fit, the elements from a followed by the elements from b. The arrays may be any length, including 0, but there will be 2 or more elements available between the 2 arrays.

i followed below approach which worked fine

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


public class TestArrays {

	/**
	 * @param args
	 */
	public static void main(String[] args) {
		// TODO Auto-generated method stub

int a[]= {4, 5};
int b[]={1, 2, 3};
System.out.println("-->"+make2( a, b)[0]+"-->"+make2( a, b)[1]);
	}



	public static int[] make2(int[] a, int[] b) {
		  List<Integer> list=new ArrayList<Integer>(a.length+b.length);
		  for( int i:a ){
		       list.add(i);
		  }
		  for( int i:b ){
		       list.add(i);
		  }
		  int[] e=new int[2];
		  e[0]=list.get(0);
		  e[1]=list.get(1);
		  return e;
		}
	
	
	
	
/*	
 * i want to use Arrays.asList approach as below but not able to use as getting compilation errors saying cannot add int to string etc. how to make below approach to work. Please advise?
 * public int[] make2(int[] a, int[] b) {
		  //int[] c=new int[2];
		  int[] d=new int[2];
		 // d=a+b;
		     List<String> list = new ArrayList<String>(Arrays.asList(a));
		    list.addAll(Arrays.asList(b));
		  int[] e=new int[2];
		  e[0]=list[0];
		  e[1]=list[1];
		  return e;
		  
		  
		  
		  
		  }*/
		  
		  
		  
}

output is
-->4-->5


when i followed above commented approach does not work. How to make above commented second approach work. Please advise

[Knute Snortum]

The code ...

int[] a = {1, 2, 3};
List<Integer> list = new ArrayList<>( Arrays.asList(a) );

will never work, because a is an array of ints (the primitive) and list is a List of Integers (the object).

for( int i:a ){
  list.add(i);
}

This works because i is being autoboxed to an Integer.

[sai rama krishna]

i wonder how to make use of asList() and addAll() approach to solve this challenge. I am fine to use Auto Boxing also if needed.

[Knute Snortum]

sai rama krishna wrote:i wonder how to make use of asList() and addAll() approach to solve this challenge. I am fine to use Auto Boxing also if needed.

Can't be done, for the reasons I give above.

[sai rama krishna]

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

public class TestArrayEx {
	public static void main(String[] args) {
		int a[] = { 4, 5 };
		int b[] = { 1, 2, 3 };

		/*
		 * Integer c[]= toObject(a); Integer d[]=toObject(b);
		 */

		System.out.println("-->" + make2(a, b)[0] + "-->" + make2(a, b)[1]);

	}

	public static int[] make2(int[] a, int[] b) {

		Integer c[] = toObject(a);
		Integer d[] = toObject(b);
		List<Integer> list = new ArrayList<Integer>(Arrays.asList(c));
		list.addAll(Arrays.asList(d));

		/*
		 * int[] result = new int[IntegerArray.length]; for (int i = 0; i <
		 * IntegerArray.length; i++) { result[i] = IntegerArray[i].intValue(); }
		 * return result;
		 */

		// Integer list2[] = new Integer[arrlist.size()];
		// list2 = arrlist.toArray(list2);

		/*
		 * int[] result = new int[list.length]; for (int i = 0; i <
		 * IntegerArray.length; i++) { result[i] = IntegerArray[i].intValue(); }
		 * return result;
		 */
		Integer list2[] = new Integer[list.size()];
		list2 = list.toArray(list2);

		int[] e = new int[2];
		e[0] = list2[0];
		e[1] = list2[1];
		return e;

	}

	public static Integer[] toObject(int[] intArray) {

		Integer[] result = new Integer[intArray.length];
		for (int i = 0; i < intArray.length; i++) {
			result[i] = Integer.valueOf(intArray[i]);
		}
		return result;
	}

}

i did as above. Not sure if it ok to approach that way. Please advise

[Knute Snortum]

Well, it's "ok", but maybe not the best way to solve the problem. I would use two for/next loops, one for each array input, and set the output array's elements until there are more than two, then exit. But looking at the program you have, here are some thoughts:

* The following code:

System.out.println("-->" + make2(a, b)[0] + "-->" + make2(a, b)[1]);

calls make2 twice, doing the same processing two times. I'd call make2 once, putting the result into an array, then print the array.

* There's no reason to put list into an array. Just use the get(index) method of java.util.List.