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Given 2 integers n and start. Your task is return any permutation p of (0,1,2.....,2^n -1) such that :

• p[0] = start
• p[i] and p[i+1] differ by only one bit in their binary representation.
• p[0] and p[2^n -1] must also differ by only one bit in their binary representation.

Example 1:

Input: n = 2, start = 3Output: [3,2,0,1]Explanation: 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]

Example 2:

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

Constraints:

• 1 <= n <= 16
• 0 <= start < 2 ^ n

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格雷码，需要知道两点：

1. 格雷码对应卡诺图，所以可以循环：

2. 格雷码的计算可以通过i^(i>>1)，其实i=range(n)，代码为

class Solution:    def circularPermutation(self, n, start):        return [start ^ i ^ (i>>1) for i in range(1<

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