1238. Circular Permutation in Binary Representation

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 = 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 <= 16
• 0 <= start < 2 ^ n

思路：一看就是格雷码

Gray code is a binary numeral system where two successive values differ in only one bit.

For example, the sequence of Gray codes for 3-bit numbers is: 000, 001, 011, 010, 110, 111, 101, 100, so G(4)=6.

下一位格雷码的计算方式：然后从后往前看，看下更新这位后得到的新数字是否在之前出现过，如果有就一直往前看，

比如起始是000，

把最后一位改掉得到001，没有出现过

然后接下来，001吧最后因为改掉成为000，出现过，所以吧001第二位改掉变成011，没出现过，所以下一个是011

以此类推，然后你可以发现G(n)=n^(n>>1)，比如第4位格雷码对应的是4^2=6

 

然后题目的意思其实就是找一个格雷码序列，然后有个start的offset而已

class Solution(object):
    def circularPermutation(self, n, start):
        """
        :type n: int
        :type start: int
        :rtype: List[int]
        """
        res = []
        for i in range(1<<n):
            res.append(start^(i^(i>>1)))
        return res
    
    
s=Solution()
print(s.circularPermutation(2,3))