(Java)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]
]

这道题和上一题主要区别在于，存在重复的数，这样，当你遍历到一个重复的数（nums[i] == nums[i-1]）的时候，若前面的数没有用过，即flags[i-1] == false，那么直接continue即可
还有这一题是未排序的，需要先行排序。
代码如下：
public class Solution {
public List<List<Integer>> permuteUnique(int[] nums) {
Arrays.sort(nums);
boolean[] flags = new boolean[nums.length];
List<List<Integer>> result = new ArrayList<List<Integer>>();
List<Integer> empty = new ArrayList<Integer>();
permuteUnique(result,empty,nums,flags,0);
return result;
}

private void permuteUnique(List<List<Integer>> result, List<Integer> empty, int[] nums, boolean[] flags, int cnt) {
// TODO Auto-generated method stub
if(cnt == nums.length){
result.add(new ArrayList<Integer>(empty));
}

for(int i = 0; i < flags.length; i++){
if(i > 0 && nums[i] == nums[i-1] && flags[i-1] == false){
continue;
}
if(flags[i] == false){
empty.add(nums[i]);
flags[i] = true;
permuteUnique(result, empty, nums, flags, cnt+1);
flags[i] = false;
empty.remove(empty.size()-1);
}

}
}
}