47. Permutations II

原题

求解带有重复元素的序列的所有排序
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]
]

思路

在无重复那道题的基础上，添加去重操作

代码

public class Solution
    {
        private IListint>> result = new Listint>>(); 
        public IListint>> PermuteUnique(int[] nums)
        {
            Array.Sort(nums);
            dfs(new List<int>(), nums, new bool[nums.Length]);
            return result;
        }

        private void dfs(IList<int> tmp, int[] nums,bool[] used)
        {
            if (tmp.Count == nums.Length)
            {
                result.Add(new List<int>(tmp));
                return;
            }
            for (int i = 0; i < nums.Length; i++)
            {
                if (used[i] == true) //深度方向递归还没完时
                continue;
                if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == false) 
                //与前一个元素相等，且前一个元素已经递归完毕
                 continue;                    
                used[i] = true;
                tmp.Add(nums[i]);
                dfs(tmp,nums,used);
                used[i] = false;
                tmp.RemoveAt(tmp.Count-1);
            }
        }
    }