1. 题目信息

给定一组不含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。

说明：解集不能包含重复的子集。

示例:输入: nums = [1,2,3]
输出:
[[3],[1],[2],[1,2,3],[1,3],[2,3],[1,2],[]
]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/subsets
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

2. 解题

2.1 暴力回溯

class Solution {
public:vector<vector<int>> subsets(vector<int>& nums) {vector<int> sub;vector<vector<int>> ans;bt(nums, sub, ans, 0);return ans;}void bt(vector<int>& nums, vector<int> sub, vector<vector<int>> & ans, int i){if(i == nums.size()){ans.push_back(sub);return;}bt(nums, sub, ans, i+1);//不加入nums[i]sub.push_back(nums[i]);bt(nums, sub, ans, i+1);//加入nums[i]}
};

or

class Solution {
public:vector<vector<int>> subsets(vector<int>& nums) {vector<int> sub;vector<vector<int>> ans;ans.push_back({});bt(nums,sub,ans,0);return ans;}void bt(vector<int>& nums,vector<int> sub, vector<vector<int>> &ans, int i){for(int idx = i; idx < nums.size(); ++idx){sub.push_back(nums[idx]);ans.push_back(sub);bt(nums,sub,ans,idx+1);sub.pop_back();}}
};

2.2 循环

class Solution {
public:vector<vector<int>> subsets(vector<int>& nums) {vector<vector<int>> ans;ans.push_back({});vector<int> sub;int i, j, n;for(i = 0; i < nums.size(); ++i){n = ans.size();//注意提前记录sizefor(j = 0; j < n; ++j){sub = ans[j];sub.push_back(nums[i]);ans.push_back(sub);}}return ans;}
};

2.3 位运算

• 对例子来讲，一共8种情况，对应的二进制位：000,001,010,011,100,101,110,111
• 每次内层循环，把为1的位对应的数字取出来，就是所有组合

class Solution {
public:vector<vector<int>> subsets(vector<int>& nums) {vector<vector<int>> ans;int i, j, k = pow(2,nums.size());for(i = 0; i < k; ++i){vector<int> sub;for(j = 0; j < nums.size(); ++j){if(i & (1<<j))sub.push_back(nums[j]);}ans.push_back(sub);}return ans;}
};