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.

解法一

这道题感觉是一个找规律的题目，找到规律后就非常好求解，感觉不是一道回溯的题。

对于n=2，它的结果包含n=1时的结果左边补零，以及逆序遍历n=1时的结果左边补1。

规律例如以下图,列出了n=1,n=2,n=3时的情况。

依据这个规律假设已知n=k的情况，那么n=k+1的结果包含对n=k的结果左边补零，即保存不变。然后逆序遍历n=k的结果左边补1就可以。

runtime:4ms

class Solution {

public:

    vector<int> grayCode(int n) {

        vector<int> result(1);

        for(int i=0;i<n;i++)

        {

                for(int j=result.size()-1;j>=0;j--)

                {

                   result.push_back((1<<i)+result[j]);

                }

        }

        return result;

    }

}。

解法二

解法二就涉及到gray code的数学知识了。要是知道这个数学知识。能够在几分钟之内就解出这道题。

格雷码能够由相应的十进制数求出：grayCode=i^i>>1

runtime:4ms

class Solution {

public:

     vector<int> grayCode(int n) {

         vector<int> result;

         for(int i=0;i<1<<n;i++)

         {

             result.push_back(i^i>>1);

         }

         return result;

     }

};