78. Subsets

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],
  []
]

backtracking

复杂度分析　　https://www.1point3acres.com/bbs/thread-117602-1-1.html

时间复杂度：还要考虑每一次要把tmp拷贝到res里的时间，复杂度是O(n)

空间复杂度：call stack - O(n)，存全部解集O(n * 2 ^ n)，O(n) << O(n * 2 ^ n)

time: O(n * 2^n), space: O(n * 2^n)

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        backtracking(nums, 0, new ArrayList<>(), res);
        return res;
    }
    
    private void backtracking(int[] nums, int idx, List<Integer> tmp, List<List<Integer>> res) {
        res.add(new ArrayList<>(tmp));
        for(int i = idx; i < nums.length; i++) {
            tmp.add(nums[i]);
            backtracking(nums, i + 1, tmp, res);
            tmp.remove(tmp.size() - 1);
        }
    }
}

time: O(2 ^ n), space: O(2 ^ n)

class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        dfs(nums, 0, new ArrayList<>(), res);
        return res;
    }
    
    private void dfs(int[] nums, int idx, List<Integer> list, List<List<Integer>> res) {
        if(idx == nums.length) {
            res.add(new ArrayList<>(list));
            return;
        }
        
        list.add(nums[idx]);
        dfs(nums, idx + 1, list, res);
        list.remove(list.size() - 1);
        
        dfs(nums, idx + 1, list, res);
    }
}