<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:
    void f(vector<int>& nList,int start,int end){
        if(start+1==end)
            return;
        if(start+2 == end){
            swap(nList[start] , nList[start+1]);
            return;
        }
        vector<int> frontList;
        vector<int> backList;
        int i = start;
        int tmp = 0;
        while(i <= end){
            if(tmp % 2==0){
                backList.push_back(nList[i]);
            }
            else{
                frontList.push_back(nList[i]);
            }
            tmp++;
            i++;
        }
        i = start;
        for(int a = 0;a < frontList.size(); ++a){
            nList[i++] = frontList[a];
        }
        for(int a = 0;a < backList.size(); ++ a){
            nList[i++] = backList[a];
        }
        int mid = (start+end)/2;
        if((start+end) % 2 == 0){
            mid--;
        }
        f(nList,start,mid);
        f(nList,mid+1,end);

    }
    vector<int> beautifulArray(int n) {
        if(n == 1) return {1};
        if(n == 2) return {1,2};
        if(n == 3) return {1,3,2};

        vector<int> resList;
        for(int i = 1;i <= n;++i){
            resList.push_back(i);
        }

        f(resList,0,n-1);
        return resList;
    }
};
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