leetcode------Gray Code

标题：	Gray Code
通过率：	32.4%
难度：	中等

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.

 

本题总体还是比较简单，就是在不知道什么是格雷码的情况下还是比较难做的，下面我贴出二进制转格雷码的实例：

按照此方法写代码，一目了然，主要就是源码前面补0的问题，然后就是如何把格雷码转换回来，如下公式：

公式表示：

（G：格雷码，B：二进制码）

java代码如下：

 1 public class Solution {

 2     public List<Integer> grayCode(int n) {

 3         List<Integer> result=new ArrayList<Integer>();

 4         String temp,gray;

 5         if(n==0){

 6             result.add(0);

 7             return result;

 8         }

 9         for(int i=0;i<Math.pow(2,n);i++){

10             temp=Integer.toBinaryString(i);

11             temp=putEquelLen(n,temp);

12             gray="";

13             for(int j=0;j<n;j++){

14                 int t=Integer.valueOf(temp.charAt(j))^Integer.valueOf(temp.charAt(j+1));

15                 gray+=t;

16             }

17             result.add(Integer.valueOf(gray,2));

18         }

19         return result;

20         

21     }

22     public String putEquelLen(int n,String str){

23        int len=str.length();

24         len=n-len+1;

25         for(int i=0;i<len;i++){

26             str="0"+str;

27         }

28         return str;

29     }

30 }

在网上发现其实就相当于一个数 (x>>1)^x的操作；

下面贴出python代码：

 1 class Solution:

 2     # @return a list of integers

 3 

 4     def grayCode(self, n):

 5         if n <= 0:

 6             return [0]

 7         ret = [0, 1]

 8         if n == 1:

 9             return ret

10 

11         for x in xrange(1, n):

12             old = list(ret)

13             new = old[::-1]

14             for (i, a) in enumerate(new):

15                 new[i] = a + (1 << x)

16             ret = old + new

17 

18         return ret