90. Subsets II

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],
  []
]

是78. Subsets的follow up。

解题思路类似，注意点首先对nums排序，然后在backtracking的时候skip duplicates。 由于每次从start位置开始，若后面有重复元素，则每次只add第一个，后面的都跳过。

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<List<Integer>>();
        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<Integer>(subset));
        if(start == nums.length) return;
        for(int i = start; i < nums.length; i++) {
            if(i != start && nums[i] == nums[i-1]) continue; // skip duplicates
            subset.add(nums[i]);
            helper(nums, i+1, subset, res);
            subset.remove(subset.size()-1);
        }
    }
}