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：

public class Solution {
    public List<Integer> grayCode(int n) {
        List<Integer> result=new ArrayList<Integer>();
        String temp,gray;
        if(n==0){
            result.add(0);
            return result;
        }
        for(int i=0;i<Math.pow(2,n);i++){
            temp=Integer.toBinaryString(i);
            temp=putEquelLen(n,temp);
            gray="";
            for(int j=0;j<n;j++){
                int t=Integer.valueOf(temp.charAt(j))^Integer.valueOf(temp.charAt(j+1));
                gray+=t;
            }
            result.add(Integer.valueOf(gray,2));
        }
        return result;
        
    }
    public String putEquelLen(int n,String str){
       int len=str.length();
        len=n-len+1;
        for(int i=0;i<len;i++){
            str="0"+str;
        }
        return str;
    }
}

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：

class Solution:
    # @return a list of integers

    def grayCode(self, n):
        if n <= 0:
            return [0]
        ret = [0, 1]
        if n == 1:
            return ret

        for x in xrange(1, n):
            old = list(ret)
            new = old[::-1]
            for (i, a) in enumerate(new):
                new[i] = a + (1 << x)
            ret = old + new

        return ret