题目描述

Given a set of distinct integers, nums, return all possible subsets (the power set).

Note: The solution set must not contain duplicate subsets.

样例

Example:

Input: nums = [1,2,3]
Output:
[
  [3],
  [1],
  [2],
  [1,2,3],
  [1,3],
  [2,3],
  [1,2],
  []
]

算法1

(dfs) $O(2^n)$

dfs,注意在往下搜索过程中，i变成i+1

C++ 代码

class Solution {
public:
    vector<vector<int>> res;
    vector<vector<int>> subsets(vector<int>& nums) {
        vector<int> vec;
        dfs(nums,0,vec);
        return res;
    }
    void dfs(vector<int>&nums,int x,vector<int>&vec){
        res.push_back(vec);//每一个都是一个集合
        for(int i=x;i<nums.size();i++){
            vec.push_back(nums[i]);
            dfs(nums,i+1,vec);
            vec.pop_back();
        }
    }
};

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度分析：blablabla

C++ 代码

blablabla