题目描述

• 不重复：用nowIndex即可，子集一定是下标有序的（相对顺序和nums[]一样）
  

代码 & 思路

• 轮流往下找即可
• 注意：通过index来实现不重复，元素相对顺序一定要和nums的一样

class Solution {public List<List<Integer>> subsets(int[] nums) {List<List<Integer>> ans = new ArrayList<>();List<Integer> nowL = new ArrayList<>();ans.add(nowL);find(0,nowL,ans,nums);return ans;}void find(int index,List<Integer> fatherL,List<List<Integer>> ans,int[] nums){// 通过index实现不重复for(int i=index;i<nums.length;i++){List<Integer> nowL = new ArrayList<>(fatherL);nowL.add(nums[i]);ans.add(nowL);find(i+1,nowL,ans,nums);}}
}

更新2.0

class Solution {List<List<Integer>> ans = new ArrayList<>();public List<List<Integer>> subsets(int[] nums) {List<Integer> list = new ArrayList<>();ans.add(list);find(nums, list, 0);return ans;}// 寻找以 index 开头的子集public void find(int[] nums, List<Integer> fatherList, int index) {for(int i = index; i < nums.length; i++) {List<Integer> list = new ArrayList<>(fatherList);list.add(nums[i]);ans.add(list);find(nums, list, i + 1);}}
}