89. Gray Code

1. 问题描述
   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 - 001 - 111 - 310 - 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.

2. 问题解决
   其实就是找规律，然后递归一下就解决了。

3. 代码

class Solution {public:    vector<int> grayCode(int n) {        if ( n == 0)            return vector<int>(1,0);        if (n == 1) {            vector<int> tmp = vector<int>(2,0);            tmp[1] = 1;            return tmp;        }        vector<int> result = grayCode(n-1);        int size = result.size();        int add = (int)pow(2.0,n-1);        for ( int i = size - 1; i >= 0; --i) {            result.push_back(result[i]+add);        }        return result;    }};

另外发现某大神的用四行代码就解决了问题，跪拜。。。

 result = [0]    for i in range(n):        result.extend([x + 2 ** i for x in result[::-1]])    return result