Given 2 integers n and start. Your task is return any permutation p of (0,1,2.....,2^n -1) such that :
p[0] = startp[i]andp[i+1]differ by only one bit in their binary representation.p[0]andp[2^n -1]must also differ by only one bit in their binary representation.
Example 1:
Input: n = 2, start = 3
Output: [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 = 2
Output: [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 <= 160 <= 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<<n)]

本文介绍了一种基于格雷码的算法,该算法能够生成一个特定的整数排列,其中任意两个相邻元素仅在二进制表示下相差一位。通过使用i^(i>>1)的巧妙计算,我们能够实现循环排列,确保了起始元素与末尾元素也仅相差一位。文章提供了Python代码示例,展示了如何生成这种特殊的排列。
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