<p>An array <code>nums</code> of length <code>n</code> is <strong>beautiful</strong> if:</p>

<ul>
	<li><code>nums</code> is a permutation of the integers in the range <code>[1, n]</code>.</li>
	<li>For every <code>0 &lt;= i &lt; j &lt; n</code>, there is no index <code>k</code> with <code>i &lt; k &lt; j</code> where <code>2 * nums[k] == nums[i] + nums[j]</code>.</li>
</ul>

<p>Given the integer <code>n</code>, return <em>any <strong>beautiful</strong> array </em><code>nums</code><em> of length </em><code>n</code>. There will be at least one valid answer for the given <code>n</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> n = 4
<strong>Output:</strong> [2,1,4,3]
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> n = 5
<strong>Output:</strong> [3,1,2,5,4]
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 1000</code></li>
</ul>

class Solution {
public:
    vector<int> beautifulArray(int n) {
        if (n == 1) return vector<int>{1};
        if (n == 2) return vector<int>{1, 2};
        int h = (n + 1) / 2;
        vector<int> a = beautifulArray(h);
        //cout << n << endl;
        vector<int> b = (n % 2 == 0)? a: beautifulArray(h - 1);
        //for (int i = 0; i < a.size(); ++i) cout << a[i];
        //cout <<h << endl;
        for (int i = 0; i < h; ++i) {
            //cout <<a[i] << endl;
            a[i] = (a[i] - 1) * 2 + 1;
            //cout <<a[i] << endl;
            if (i < b.size()) b[i] = b[i] * 2;
        }
        //cout <<a[h - 1] << endl;
        //cout <<a[h - 1] << endl;
        a.insert(a.end(), b.begin(), b.end());
        //for (int i = 0; i < a.size(); ++i) cout << a[i];
        //cout <<endl;
        return a;
    }
};
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