LeetCode-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.

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

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

使用回溯法，具体可参考Combination Sum中的讲解。

针对这题我们需要求出给定集合的所有子集，所以当然也包括空集。以[1,2,3]为例，搜索的策略：

1. 先把空集加入到返回链表；
2. 再有下图的顺序搜索得到所有的解。
   

class Solution {    public List<List<Integer>> subsets(int[] nums) {        List<List<Integer>> res=new LinkedList<List<Integer>>();        subsetsBacktrack(res,new LinkedList<Integer>(),nums,0);        return res;    }    private void subsetsBacktrack(List<List<Integer>> subsetsList,LinkedList<Integer> tempList,int[] nums,int start)    {        subsetsList.add(new LinkedList<Integer>(tempList));        for(int i=start;i<nums.length;++i)        {            tempList.add(nums[i]);            subsetsBacktrack(subsetsList,tempList,nums,i+1);            tempList.remove(tempList.size()-1);        }    }}