<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
{
public:
    vector<vector<int>> permute(vector<int>& nums)
    {
        int n = nums.size();
        int size = n;
        while (--n > 1) size *= n;
        n = nums.size();

        vector<vector<int>> res;
        res.resize(size);
        int r = 0;

        vector<int> temp;
        temp.resize(n);

        vector<int> idxs;
        idxs.resize(n);

        vector<int> used;
        used.resize(n);

        int i = 0;
        idxs[0] = 0;
        while (true)
        {
            if (i < n-1)
            {
                if (idxs[i] < n - i)
                {
                    idxs[++i] = 0;
                }
                else
                {
                    if (i == 0) break;

                    idxs[--i] += 1;
                }
            }
            else
            {
                for (int j = 0; j < n; j++)
                    used[j] = false;

                for (int j = 0; j < n; j++)
                {
                    int k = 0;
                    for (int c = 0; c <= idxs[j]; c++)
                    {
                        while (used[k] == true) k++;
                        k++;
                    }
                    k--;
                    used[k] = true;
                    temp[j] = nums[k];
                }
                
                res[r++] = temp;

                if (i == 0) break;

                idxs[--i] += 1;
            }
        }

        return res;
    }
};
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