78. Subsets

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

Note: The solution set must not contain duplicate subsets.

For example,
If nums = [1,2,3], a solution is:

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

一刷
题解：
back tracking
需要补充的点：怎么求时间复杂度，还有怎么用bit manipulation

public class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> res = new ArrayList<>();
        if(nums == null || nums.length == 0) return res;
        boolean[] visited = new boolean[nums.length];
        subsets(res, nums, new ArrayList<Integer>(), visited, 0);
        return res;
    }
    
    private void subsets(List<List<Integer>> res, int[] nums, List<Integer> list, boolean[] visited, int pos){
        res.add(new ArrayList<Integer>(list));
        
        for(int i=pos; i<nums.length; i++){
            if(!visited[i]){
                list.add(nums[i]);
                visited[i] = true;
                subsets(res, nums, list, visited, i+1);
                list.remove(list.size()-1);
                visited[i] = false;
            }
        }
    }
}

二刷:

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