Permutations II(lintcode 16)

Description

Given a list of numbers with duplicate number in it. Find all unique permutations.

Example

For numbers [1,2,2] the unique permutations are:

[
  [1,2,2],
  [2,1,2],
  [2,2,1]
]

Interface

class Solution {
    /**
     * @param nums: A list of integers.
     * @return: A list of unique permutations.
     */
    public List<List<Integer>> permuteUnique(int[] nums) {
        // Write your code here
    } 
}

Solution

class Solution {
    /**
     * @param nums: A list of integers.
     * @return: A list of unique permutations.
     */
    public List<List<Integer>> permuteUnique(int[] nums) {
        // Write your code here
        List<List<Integer>> result = new ArrayList<List<Integer>>();

        if (nums == null) {
            return result;
        }

        if (nums.length == 0) {
            result.add(new ArrayList<Integer>());
            return result;
        }

        List<Integer> list = new ArrayList<Integer>();
        Arrays.sort(nums);
        int[] isUsed = new int[nums.length];
        for (int i = 0; i < nums.length; ++i) {
            isUsed[i] = 0;
        }

        helper(result, list, isUsed, nums);
        return result;
    } 

    private void helper(List<List<Integer>> result, List<Integer> list, int[] isUsed, int[] nums) {
        if (list.size() == nums.length) {
            result.add(new ArrayList<Integer>(list));
            return;
        }

        for (int i = 0; i < nums.length; ++i) {
            if (isUsed[i] == 1 || (i != 0 && nums[i] == nums[i - 1] && isUsed[i - 1] == 0)) {
                continue;
            }

            isUsed[i] = 1;
            list.add(nums[i]);
            helper(result, list, isUsed, nums);
            list.remove(list.size() - 1);
            isUsed[i] = 0;
        }
    }
}