Java for LeetCode 089 Gray Code

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.

解题思路：

格雷码，学过通信原理的童鞋应该都不陌生，只不过格雷码的生成方法不是按照题目中描述的那样，下面科普下格雷码的生成方法：

n=1 得到0 1

n=2 得到00 01 11 10（注意，11 和 10除了符号位外，和 00 01对称的）

n=3 得到000 001 011 010 110 111 101 100（后面四位除了符号位之外，和前四位对称）

n=4 依次类推

知道了格雷码的生成过程，那么JAVA实现如下：

   static public List<Integer> grayCode(int n) {
        List<Integer> list=new ArrayList<Integer>();
        for(int i=0;i<Math.pow(2, n);i++){
        	int result=0,num=i,temp=n;
        	while(temp>1){
        		if(num>=Math.pow(2, temp-1)){
        			num=(int) (2*Math.pow(2, temp-1)-num-1);
        			result+=Math.pow(2, temp-1);
        		}
        		temp--;
        	}
        	result+=num;
        	list.add(result);
        }
        return list;
    }

 同时，本题有一个非常简洁的写法，如下：

    public List<Integer> grayCode(int n) {
        List<Integer> res = new ArrayList<Integer>();
        for (int i = 0; i < (1<<n); ++i)
            res.add(i ^ (i>>1));
        return res;
    }