Subset II

Leetcode

題意：

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

Note: Elements in a subset must be in non-descending order. 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],
  []
]

解題思路：

如果有重複的話，不能跳著取。

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();

        if (nums == null || nums.length == 0) {
            return res;   
        }

        Arrays.sort(nums);

        helper(nums, 0, new ArrayList<Integer>(), res);
        return res;
    }

    private void helper(int[] nums, int pos, List<Integer> tempRes, List<List<Integer>> res) {
        if (pos > nums.length) {
            return;
        }

        res.add(new ArrayList<Integer>(tempRes));
        for (int i = pos; i < nums.length; i++) {
            if (i == pos || nums[i - 1] != nums[i]) {
                tempRes.add(nums[i]);
                helper(nums, i + 1, tempRes, res);
                tempRes.remove(tempRes.size() - 1);
            }
        }
    }
}

Bit Manipulation:

public class Solution {
    public List<List<Integer>> subsetsWithDup(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();

        if (nums == null || nums.length == 0) {
            return res;   
        }

        Arrays.sort(nums);

        int powerOfTwo = (int)Math.pow(2, nums.length);
        for (int i = 0; i < powerOfTwo; i++) {
            List<Integer> list = new ArrayList<>();
            boolean isIllegal = false;
            for (int j = 0; j < nums.length; j++) {
                if (j > 0 && (nums[j] == nums[j - 1]) && ((i >> (j - 1)) & 1) == 0) {
                    isIllegal = true;
                    break;
                } else {
                    list.add(nums[j]);
                }
            }
            if (isIllegal == false) {
                res.add(list);
            }

        }

        return res;
    }


}