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].

题目大意：给一个数组，包含重复元素，返回所有可能组合，去重。

解题思路：这题是Permutation的变种，思路就是回溯。首先把数组排序一下，对于不是第一次使用的数字，检测是否出现过。

    public List<List<Integer>> permuteUnique(int[] nums) {

        List<List<Integer>> res = new ArrayList<>();

        if (nums == null || nums.length == 0) {

            return res;

        }

        Arrays.sort(nums);

        List<Integer> tmp = new ArrayList<>();

        boolean[] used = new boolean[nums.length];

        helper(res, tmp, 0, nums.length, nums, used);

        return res;

    }



    public void helper(List<List<Integer>> res, List<Integer> tmp, int n, int N, int[] nums, boolean[] used) {

        if (n == N) {

            res.add(new ArrayList<>(tmp));

            return;

        }

        int last_num = 0x80000000;

        for (int i = 0; i < N; i++) {

            if (!used[i] && last_num != nums[i]) {

                used[i] = true;

                tmp.add(nums[i]);

                last_num = nums[i];

                helper(res, tmp, n + 1, N, nums, used);

                used[i] = false;

                tmp.remove(tmp.size() - 1);

            }

        }

    }