LeetCode78

LeetCode78

这道题比较简单~一次就AC了~

但是我觉得求求集合子集的算法是一个很重要的子算法，还是在这里记录一下自己的思路。

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:

我的思路是类似二叉树穷举，在每层分叉时进行两次递归，一次加入待选元素，一次不加入。

public class Solution {public List<List<Integer>> subsets(int[] nums) {List<List<Integer>> retlist = new ArrayList<List<Integer>>();List<Integer> cur = new ArrayList<Integer>();tree(retlist,cur,0,nums);return retlist;}private void tree (List<List<Integer>> retlist,List<Integer> cur,int level,int[] nums){List<Integer> curnew = new ArrayList<Integer>();curnew.addAll(cur);if (level == nums.length){retlist.add(curnew);return;}tree(retlist,curnew,level+1,nums);curnew.add(nums[level]);tree(retlist,curnew,level+1,nums);}
}