78.& 90. Subsets I & II

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

Example:

Input: nums = [1,2,3]
Output:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

代码：

public List<List<Integer>> subsets(int[] nums) {
    if (nums == null) {
        return null;
    }
    List<List<Integer>> res = new ArrayList<>();
    ArrayList<Integer> emptyList = new ArrayList<>();
    res.add(emptyList);

    for(int i = 0; i < nums.length; i++) {
        int size = res.size();
        for(int j = 0; j < size; j++) {
            ArrayList<Integer> newList = new ArrayList<Integer>(res.get(j));
            newList.add(nums[i]);
            res.add(newList);
        }
    }

    return res;
}

Follow Up: 90. Subsets II

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],
  []
]

Solution:

public List<List<Integer>> subsetsWithDup(int[] nums) {
    List<List<Integer>> res = new ArrayList<>();
    if (nums == null) {
        return res;
    }
    Arrays.sort(nums);
    ArrayList<Integer> emptyList = new ArrayList<>();
    res.add(emptyList);

    int sameCount = 1;
    for(int i = 0; i < nums.length; i++) {
        if (i > 0 && nums[i] == nums[i - 1]) {
            sameCount++;
        } else {
            sameCount = 1;
        }

        int size = res.size();
        int start = size - size / sameCount;
        for (int j = start; j < size; j++) {
            ArrayList<Integer> newList = new ArrayList<Integer>(res.get(j));
            newList.add(nums[i]);
            res.add(newList);
        }
    }

    return res;
}