【leetcode】Gray Code

题目：

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.
思路：
这是一道非常有意思的题。初看无下手之处，不过列举出前几项就很容易看出规律啦：

n = 1时：

0 （0）

1 （1）

n = 2时：

00 (0)

01 (1)

11 (3)

10 (2)

n = 3时：

000 (0)

001 (1)

011 (3)

010 (2)

110 (6)

111 (7)

101 (5)

100 (4)

可看出n每增加一位，结果数量翻倍，且前一半和n-1的结果是一样的，而后一半其实就是前半部分的镜像对称，再在最高位加上1.

所以完全可以用递归，通过n-1得到n的结果。代码如下：

public List<Integer> grayCode(int n) {
List<Integer> result = new ArrayList<Integer>();
if(n<=1) {
for(int i =0; i<=n; i++) {
result.add(i);
}
return result;
}
result = grayCode(n-1);
List<Integer> r1 = reverse(result);
int x = 1<<(n-1);
for(int i = 0; i< r1.size(); i++) {
r1.set(i, r1.get(i)+x);
}
result.addAll(r1);
return result;
}

public List<Integer> reverse(ArrayList<Integer> r) {
List<Integer> result = new ArrayList<Integer>();
for(int i = r.size()-1; i>=0; i--) {
result.add(r.get(i));
}
return result;
}