<p>Given an array <code>nums</code> of distinct integers, return <em>all the possible permutations</em>. You can return the answer 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> [[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> nums = [0,1]
<strong>Output:</strong> [[0,1],[1,0]]
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> nums = [1]
<strong>Output:</strong> [[1]]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 6</code></li>
	<li><code>-10 &lt;= nums[i] &lt;= 10</code></li>
	<li>All the integers of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
private:
    void swap(int & a, int & b){
        int t = a;
        a = b;
        b = t;
    }
    void permute(vector<vector<int>> & permutations, vector<int> nums, int l, int r){
        if(l >= r){ 
            permutations.push_back(nums);
            return;
        }
        for(int i=l; i<=r; i++){
            swap(nums[i], nums[l]);
            permute(permutations, nums, l+1, r);
            swap(nums[i], nums[l]);
        }
    }
public:
    vector<vector<int>> permute(vector<int>& nums) {
        int n = nums.size();

        vector<vector<int>> res;

        permute(res, nums, 0, n-1);

        return res;
    }
};

// a = 1000
// b = 0111

// a = 1111
// b = 1000
// a = 0111

// 0^0 = 0
// 1^0 = 1
// 0^1 = 1
// 1^1 = 0

// 01 ^ 11 = 10
// 10 ^ 11 = 01
// 10 ^ 01 = 11
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