Permutations II @LeetCode

Given a collection of numbers that might contain duplicates, return all possible unique permutations.

For example,
[1,1,2] have the following unique permutations:
[1,1,2], [1,2,1], and [2,1,1].

	package leetcode.combinations;
	import java.util.ArrayList;
	import java.util.Arrays;
	/**
	* Thought: Almost same as Permutation. Only difference is to remove duplicates.
	* if number1 equals number2, defaultly, we get the number1, abort number2, so
	* number2 can't be used if the number1 haven't be used.
	*
	* In subset, we have a position marker and sorted int array. As we don't have position now, we could
	* use a flag array instead.
	*
	* @author jeffwan
	* @date Feb 10, 2014
	*/
	public class Permutations2 {
	public static void main(String[] args) {
	int[] num = {1, 1, 2};
	System.out.println(permuteUnique(num));
	}
	public static ArrayList<ArrayList<Integer>> permuteUnique(int[] num) {
	Arrays.sort(num);
	ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
	ArrayList<Integer> list = new ArrayList<Integer>();
	boolean[] visited = new boolean[num.length];
	permuteUniqueHelper(result, list, num, visited);
	return result;
	}
	private static void permuteUniqueHelper(ArrayList<ArrayList<Integer>> result,
	ArrayList<Integer> list, int[] num, boolean[] visited) {
	if (list.size() == num.length) {
	result.add(new ArrayList<Integer>(list));
	return;
	}
	for (int i=0; i<num.length; i++) {
	if (visited[i]==true || (i!=0 && num[i] == num[i-1] && visited[i-1] == false)) {
	continue;
	}
	visited[i] = true;
	list.add(num[i]);
	permuteUniqueHelper(result, list, num, visited);
	list.remove(list.size() - 1);
	visited[i] = false;
	}
	}
	}

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