[leetcode] 47. Permutations II

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],
  [2,1,1]
]
 
和一般的Permutation不一样的是，这种permutation需要排序，使相同的元素能够相邻，选取下一个元素的时候，要查看这个元素的前一个元素是否和它相同，如果相同而且没有使用过，就不用选取这个元素，因为如果选取了这个元素，所得的结果被包含于选取了前一个相同元素的结果中。
代码如下：

public class Solution {
    int length;
    public List<List<Integer>> permuteUnique(int[] nums) {
        length = nums.length;
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (length == 0) {
            return result;
        }
        Arrays.sort(nums);
        boolean[] flags = new boolean[length];
        Arrays.fill(flags, false);
        List<Integer> candidate = new ArrayList<Integer>();
        helper(nums, length, flags, result, candidate);
        return result;
    }
    private void helper(int[] nums, int n, boolean[] flags, List<List<Integer>> result, List<Integer> candidate) {
        if (n == 0) {
            result.add(new ArrayList<Integer>(candidate));
        }
        int ll = candidate.size();
        for (int i = 0; i < length; i++) {
            if (!flags[i]) {
                //if the number appears more than once and the same number before it has not been used
                if (i != 0 && nums[i] == nums[i - 1] && !flags[i - 1]) {
                    continue;
                }
                flags[i] = true;
                candidate.add(nums[i]);
                helper(nums, n - 1, flags, result, candidate);
                candidate.remove(ll);
                flags[i] = false;
            }
        }
    }
}