<p>Given an integer array <code>nums</code> of <strong>unique</strong> elements, return <em>all possible</em> <span data-keyword="subset"><em>subsets</em></span> <em>(the power set)</em>.</p>

<p>The solution set <strong>must not</strong> contain duplicate subsets. Return the solution in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3]
<strong>Output:</strong> [[],[1],[2],[1,2],[3],[1,3],[2,3],[1,2,3]]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [0]
<strong>Output:</strong> [[],[0]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
	<li>All the numbers of&nbsp;<code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    vector<vector<int>> output;
    int n, k;
    void backtrack(int first, vector<int> curr, vector<int>& nums) {
        // if the combination is done
        if (curr.size() == k) output.push_back(curr);
        for (int i = first; i < n; ++i) {
            // add i into the current combination
            curr.push_back(nums[i]);
            // use next integers to complete the combination
            backtrack(i + 1, curr, nums);
            // backtrack
            curr.pop_back();
        }
    }
    vector<vector<int>> subsets(vector<int>& nums) {
        n = nums.size();
        for (k = 0; k < n + 1; ++k) {
            vector<int> curr;
            backtrack(0, curr, nums);
        }
        return output;
    }
};
10700