[Algorithm] 九章一：算法面试与代码风格

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

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

Example

If S = [1,2,3], a solution is:

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

Challenge ：Can you do it in both recursively and iteratively?

思路：dfs

public class Solution {    public List<List<Integer>> subsets(int[] nums) {        List<List<Integer>> results=new ArrayList<>();        if(nums==null){            return results;        }                if(nums.length==0){            results.add(new ArrayList<Integer>());            return results;        }        Arrays.sort(nums);        helper(new ArrayList<Integer>(),nums,0,results);        return results;    }        private void helper(ArrayList<Integer> subset,int [] nums,int startIndex,List<List<Integer>> results){        results.add(new ArrayList<>(subset));         for(int i=startIndex;i<nums.length;i++){                  subset.add(nums[i]);                  //例如[]->[1]            helper(subset,nums,i+1,results);      //寻找所有以1开头的子集，并加入到results里，再同样做子任务            subset.remove(subset.size()-1);        }    }}