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.

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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;

    }

}。

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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;

     }

};