# 78.subsets

## Statement

• Difficulty: Medium
• Tag: `位运算` `数组` `回溯`

``````输入：nums = [1,2,3]

``````

``````输入：nums = [0]

``````

• `1 <= nums.length <= 10`
• `-10 <= nums[i] <= 10`
• `nums` 中的所有元素 互不相同

• Difficulty: Medium
• Tag: `Bit Manipulation` `Array` `Backtracking`

Given an integer array `nums` of unique elements, return all possible subsets (the power set).

The solution set must not contain duplicate subsets. Return the solution in any order.

Example 1:

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

Example 2:

``````Input: nums = [0]
Output: [[],[0]]
``````

Constraints:

• `1 <= nums.length <= 10`
• `-10 <= nums[i] <= 10`
• All the numbers of `nums` are unique.

## Solution

``````#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>

#define endl "\n"
#define fi first
#define se second
#define all(x) begin(x), end(x)
#define rall rbegin(a), rend(a)
#define bitcnt(x) (__builtin_popcountll(x))
#define complete_unique(a) a.erase(unique(begin(a), end(a)), end(a))
#define mst(x, a) memset(x, a, sizeof(x))
#define MP make_pair

using ll = long long;
using ull = unsigned long long;
using db = double;
using ld = long double;
using VLL = std::vector<ll>;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using PLL = std::pair<ll, ll>;

using namespace __gnu_pbds;
using namespace std;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
const ll mod = 1e9 + 7;

template <typename T, typename S>
inline bool chmax(T &a, const S &b) {
return a < b ? a = b, 1 : 0;
}

template <typename T, typename S>
inline bool chmin(T &a, const S &b) {
return a > b ? a = b, 1 : 0;
}

#ifdef LOCAL
#include <debug.hpp>
#else
#define dbg(...)
#endif

class Solution {
public:
vector<vector<int>> subsets(vector<int> &nums) {
int n = nums.size();
auto res = vector<vector<int>>();

for (int S = 0; S < (1 << n); S++) {
auto cur = vector<int>();
for (int i = 0; i < n; i++) {
if ((S >> i) & 1) {
cur.push_back(nums[i]);
}
}
res.push_back(cur);
}

return res;
}
};

#ifdef LOCAL

int main() {
return 0;
}

#endif
``````