Code puzzler: Tape Equilibrium

Suppose you get an array of ints of arbitrary length. Now we need to find out what is the mimimal absolute difference of the 2 sums of numbers if you split the array at any position in a left and right array. And preferably in a efficient way.

	import java.util.ArrayList;
	import java.util.List;
	public class TapeEquilibrium {
	public static void main(String[] args) {
	int [] numbers1 = {1,3,2,7};
	int [] numbers2 = {5,2,8,4};
	evaluate(numbers1, 1);
	evaluate(numbers2, 5);
	}
	public static void evaluate(int [] numbers, int expected) {
	int answer = solution(numbers);
	System.out.println("Expected answer = " + expected + ", got " + answer);
	}
	public static int solution(int[] A) {
	List<Integer> numbers = toList(A);
	int size = numbers.size();
	int sumLeft = numbers.get(0);
	int sumRight = sum(numbers.subList(1, size));
	int minDifference = Math.abs(sumLeft - sumRight);
	for (int i : numbers.subList(1, size -1)) {
	sumLeft = sumLeft + i;
	sumRight = sumRight - i;
	int difference = Math.abs(sumLeft - sumRight);
	if (difference < minDifference) {
	minDifference = difference;
	}
	}
	return minDifference;
	}
	public static List<Integer> toList(int [] array) {
	List<Integer> result = new ArrayList<Integer>();
	for (int ele : array) {
	result.add(new Integer(ele));
	}
	return result;
	}
	public static int sum(List<Integer> numbers) {
	int sumRight = 0;
	for (int i : numbers) {
	sumRight += i;
	}
	return sumRight;
	}
	}

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