生成格雷码（Gray Code）

题目原型：

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.

基本思路：

首先，当n=1时，格雷码为{“0”}，{“1”}，当n=2时，格雷码为{“00”，“01”}，{“11”，“10”}，当n=3时，格雷码为{“000”，“001”，“011”，“010”}，{“110”，“111”，“101”，“100”}。。。看到这里我们是否发现了一个规律，就

是n位的格雷码是在n-1的基础上，先从前往后扫描，在最前面插入一个“0”，然后从后往前扫描，插入一个“1”.

以此类推，便可得到结果。

public ArrayList<Integer> grayCode(int n) {ArrayList<Integer> answer = new ArrayList<Integer>();if(n==0){answer.add(0);return answer;}ArrayList<String> result = generateCode(n);for(String str : result){answer.add(transfer(str));}return answer;    }//二进制转十进制public int transfer(String binary){int result = 0;for(int i = binary.length()-1;i>=0;i--){result+=(binary.charAt(i)-'0')*(int)Math.pow(2, binary.length()-i-1);}return result;}public ArrayList<String> generateCode(int n){ArrayList<String> tmp = new ArrayList<String>();ArrayList<String> result = new ArrayList<String>();tmp.add("0");tmp.add("1");if(n==1){return tmp;}for(int i = 2;i<=n;i++){result = new ArrayList<String>();for(String str : tmp){result.add("0"+str);}for(int j = tmp.size()-1;j>=0;j--){result.add("1"+tmp.get(j));}tmp = result;}return result;}