题目：（Backtracking）

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.

题解：

参考引用http://www.lifeincode.net/programming/leetcode-gray-code-java/

We can see that the last two digits of 4 codes at the bottom is just the descending sequence of the first 4 codes. The first 4 codes are 0, 1, 3, 2. So, we can easily get the last 4 codes: 2 + 4, 3 + 4, 1 + 4, 0 + 4, which is 6, 7, 5, 4. We can keep doing this until we reach n digits.

要解决这个问题先看一下greycode的规律

当n=1：

0

1

当n=2：

00

01

11

10

当 n = 3:

000
001
011
010
110
111
101
100

然后我们可以发现一个规律，当n每增加1的时候，

其实就是：1.把n-1的所有元素前面加上一个0.  2.将n-1的所有元素倒序排列以后前面加上1。

例如当 n=2时 n-1 为 

0

1

所有第一步将n-1的所有元素前面加上一个0

00

01

第二步将n-1的所有元素倒序排列以后前面加上1

11

10

然后因为要转化成十进制的数，第一步前面加0，就是加0，就是什么都不要加。第二步前面要加1，正好就是加上 n-1 的所有元素的 数目。

所以代码如下：

public class Solution {    public List<Integer> grayCode(int n) {        List<Integer> ret = new LinkedList<>();        ret.add(0);        for (int i = 0; i < n; i++) {            int size = ret.size();            for (int j = size - 1; j >= 0; j--)                ret.add(ret.get(j) + size);        }        return ret;    }}

[leetcode] Grey Code