> For the complete documentation index, see [llms.txt](https://aaronice.gitbook.io/lintcode/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://aaronice.gitbook.io/lintcode/backtracking/subsets-ii.md).

# Subsets II

Medium

Given a collection of integers that might contain duplicates, ***nums***, return all possible subsets (the power set).

**Note:**The solution set must not contain duplicate subsets.

**Example:**

```
Input:
 [1,2,2]

Output:

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

## Analysis

这道题应该算Subsets的follow-up，解决input中有重复元素的情况。

不过对策也很直观，就是当遇到重复元素时，跳过相应的backtracking即可。

## Solution

Backtracking

```java
class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        Arrays.sort(nums);
        helper(nums, 0, new ArrayList<Integer>(), res);
        return res;
    }
    private void helper(int[] nums, int start, List<Integer> subset, List<List<Integer>> res) {
        res.add(new ArrayList(subset));
        for (int i = start; i < nums.length; i++) {
            if (i > start && nums[i] == nums[i - 1]) {
                continue;
            }
            subset.add(nums[i]);
            helper(nums, i + 1, subset, res);
            subset.remove(subset.size() - 1);
        }
    }
}
```

Iterative

```java
public List<List<Integer>> subsetsWithDup(int[] nums) {
    Arrays.sort(nums);
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    result.add(new ArrayList<Integer>());
    int begin = 0;
    for(int i = 0; i < nums.length; i++){
        if(i == 0 || nums[i] != nums[i - 1]) begin = 0;
        int size = result.size();
        for(int j = begin; j < size; j++){
            List<Integer> cur = new ArrayList<Integer>(result.get(j));
            cur.add(nums[i]);
            result.add(cur);
        }
        begin = size;
    }
    return result;
}
```


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