<p>Given 2 integers <code>n</code> and <code>start</code>. Your task is return <strong>any</strong> permutation <code>p</code>&nbsp;of <code>(0,1,2.....,2^n -1) </code>such that :</p>

<ul>
	<li><code>p[0] = start</code></li>
	<li><code>p[i]</code> and <code>p[i+1]</code>&nbsp;differ by only one bit in their binary representation.</li>
	<li><code>p[0]</code> and <code>p[2^n -1]</code>&nbsp;must also differ by only one bit in their binary representation.</li>
</ul>

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

<pre>
<strong>Input:</strong> n = 2, start = 3
<strong>Output:</strong> [3,2,0,1]
<strong>Explanation:</strong> The binary representation of the permutation is (11,10,00,01). 
All the adjacent element differ by one bit. Another valid permutation is [3,1,0,2]
</pre>

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

<pre>
<strong>Input:</strong> n = 3, start = 2
<strong>Output:</strong> [2,6,7,5,4,0,1,3]
<strong>Explanation:</strong> The binary representation of the permutation is (010,110,111,101,100,000,001,011).
</pre>

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

<ul>
	<li><code>1 &lt;= n &lt;= 16</code></li>
	<li><code>0 &lt;= start&nbsp;&lt;&nbsp;2 ^ n</code></li>
</ul>
class Solution {
public:
    vector<int> circularPermutation(int n, int start) {
        vector<int> graycode, answer;
        graycode.push_back(0);
        graycode.push_back(1);
        vector<int> rev;
        int offset;
        for (int power = 1; power<n; power++){
            rev=graycode;
            reverse(rev.begin(), rev.end());
            graycode.insert(graycode.end(), rev.begin(), rev.end());
            offset=1<<power;
            for (int index=offset; index<(offset*2); index++){
                graycode[index]+=(offset);
            }
        }

        vector<int>::iterator iter=graycode.begin();
        for (iter; iter<graycode.end(); iter++){
            if((*iter)==start){
                answer.insert(answer.begin(), iter, graycode.end());
                answer.insert(answer.end(), graycode.begin(), iter);
                break;
            }
        }
        return answer;
    }
};
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