90. Subsets II

题目

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,2], a solution is:

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

代码

不含重复元素的解法：
http://blog.csdn.net/daigualu/article/details/76348947
与解决不含重复元素的子集思路一致，只有两处不同：
1. 需要对数组排序
2. 过滤重复项

public class Solution {    private IList<IList<int>> rtn = new List<IList<int>>();       public IList<IList<int>> SubsetsWithDup(int[] nums)        {            Array.Sort(nums); //diff1: in-place sort            rtn.Add(new List<int>());            rtn.Add(new List<int>(nums));            dfs2(nums, new List<int>(), 0, 1); //call dfs            return rtn;        }        //assert: ndigts>=1        //dfs: two recursion return conditions for dfs :        //1. size for an item of this subsets increases to ndigits        //2. depth-first-search for nums[i] with ndigits is to the end point.         private void dfs2(int[] nums, IList<int> item, int start, int ndigts)        {            if (ndigts == item.Count)            {                rtn.Add(new List<int>(item));                return; //dfs recursion return condition 1            }            while (ndigts < nums.Length)            {                for (int i = start; i < nums.Length; i++)                {                    //diff2: filter duplicates                    if (i > start && nums[i] == nums[i - 1])                         continue;                    item.Add(nums[i]);                    dfs2(nums, item, i + 1, ndigts); //forward 1 to search                    item.RemoveAt(item.Count - 1);                }                if (item.Count == 0)                    ndigts++;                else //dfs recursion return condition 2                    return;            }        }}