Problem

Given a set of distinct integers, return all possible subsets.

 Notice

Elements in a subset must be in non-descending order.
The solution set must not contain duplicate subsets.
Have you met this question in a real interview? Yes
Example
If S = [1,2,3], a solution is:

[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

思路

1. 假设数组: [1, 2, 3]
2. 找到整个数组的subsets相当于分别找到以1, 2, 3开头的subset
3. 找到1开头的subset, 相当于1已经存在于路径当中, 找到1后面, 即分别以2, 3 开头的subset
4. 直到数组遍历完为止

两种写法

    public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();
        result.add(new ArrayList<Integer>());
        if (nums == null || nums.length == 0) return result;
        Arrays.sort(nums);
        helper(nums, result, new ArrayList<Integer>(), 0);
        return result;
    }

    private void helper(int[] nums, ArrayList<ArrayList<Integer>> result, ArrayList<Integer> path, int startIndex) {
        if (nums.length == startIndex) return;
        for (int i = startIndex; i < nums.length; i++) {
            path.add(nums[i]);
            result.add(new ArrayList<Integer>(path));
            helper(nums, result, path, i + 1);
            path.remove(path.size() - 1);
        }
    }

    public ArrayList<ArrayList<Integer>> subsets(int[] nums) {
        ArrayList<ArrayList<Integer>> result = new ArrayList<>();
        if (nums == null || nums.length == 0) {
            result.add(new ArrayList<Integer>());
            return result;
        }
        Arrays.sort(nums);
        helper(nums, result, new ArrayList<Integer>(), 0);
        return result;
    }

    private void helper(int[] nums, ArrayList<ArrayList<Integer>> result, ArrayList<Integer> path, int startIndex) {
        if (nums.length < startIndex) return;
        result.add(new ArrayList<Integer>(path));
        for (int i = startIndex; i < nums.length; i++) {
            path.add(nums[i]);
            helper(nums, result, path, i + 1);
            path.remove(path.size() - 1);
        }
    }