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题目描述： The gray code is a binary numeral system where two successive values differ in only one bit. Given a non-negative integer n representing the total number of bits in the code, print the sequence of gray code.  A gray code sequence must begin with 0. For example, given n = 2, return [0,1,3,2]. Its gray code sequence is: 00 - 0 01 - 1 11 - 3 10 - 2 Note: For a given n, a gray code sequence is not uniquely defined.

For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.

For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.

分析：

n=3时的格雷码如下：

000

001 011 010

----------对称轴

110

111 101 100 前4个最高位是0，而后4个最高位是1 将上述格雷码的最高位去掉，观察剩下的2位：前4个恰好是n=2的格雷码，而后4个是前4个的逆序。 因此，我们将格雷码看成是上下两部分，如下图： 上半部分是n=2的格雷码（最高位多了一个0，但这对结果并没有影响）； 下半部分是n=2的格雷码的逆序，然后在最高位加1（本例中，最高位加1 等价于 将格雷码加4）

python代码1：

class Solution(object):    def grayCode(self, n):        """        :type n: int        :rtype: List[int]        """        if n==0:            return [0]        self.sequence=[0,1]        if n==1:            return self.sequence        for i in [2**x for x in range(1,n)]:            self.sequence.extend([i+v for v in self.sequence[::-1]])            #或者使用sequence[None:None:-1]，表示通过切片获得列表sequence的反向副本，切片操作不改变列表sequence本身        return self.sequence

Python代码2：

class Solution(object):    def grayCode(self, n):        """        :type n: int        :rtype: List[int]        """        self.sequence=[0]        for i in [1<

C++代码：

class Solution {public:    vector
   
     grayCode(int n) {        vector
    
     sequence;        sequence.push_back(0);        for(int i=0;i
     
      =0;i--)            {                sequence.push_back(highest+sequence[i]);            }        }        return sequence;    }};