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.

题解

这道题是要考我知不知道格雷码么。。反正我自己单看找不出规律。。。

可以看到n位的格雷码由两部分构成,一部分是n-1位格雷码,再加上 1<<(n-1)和n-1位格雷码的逆序(整个格雷码逆序0132变成2310这种)的和。

1位格雷码有两个码字
(n+1)位格雷码中的前2^n个码字等于n位格雷码的码字,按顺序书写,加前缀0
(n+1)位格雷码中的后2^n个码字等于n位格雷码的码字,按逆序书写,加前缀1。

由于是二进制,在最高位加0跟原来的数本质没有改变,所以取得上一位算出的格雷码结果,再加上逆序添1的方法就是当前这位格雷码的结果了。

n = 0时,[0]

n = 1时,[0,1]

n = 2时,[00,01,11,10]

n = 3时,[000,001,011,010,110,111,101,100]

当n=1时,0,1

当n=2时,原来的list 0,1不变,只是前面形式上加了个0变成00,01。然后加数是1<<1为10,依次:10+1=11 10+0=10。结果为:00 01 11 10

当n=3时,原来的list 00,01,11, 10(倒序为:10,11,01,00)。加数1<<2为100。倒序相加为:100+10=110, 100+11=,100+01=, 100+00= 100。

最终结果为000 001 011 010 110 111 101 100

代码如下:

 1     public ArrayList<Integer> grayCode(int n) {  
 2         if(n==0) {  
 3             ArrayList<Integer> result = new ArrayList<Integer>();  
 4             result.add(0);  
 5             return result;  
 6         }  
 7           
 8         ArrayList<Integer> result = grayCode(n-1);  
 9         int addNumber = 1 << (n-1);
         int originalsize=result.size();
         
         for(int i=originalsize-1;i>=0;i--) {  
             result.add(addNumber + result.get(i));  
         }  
         return result;  
     }

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