Gray Code leetcode java
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=111,100+01=101, 100+00= 100。
最终结果为000 001 011 010 110 111 101 100
代码如下:
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);
10 int originalsize=result.size();
11
12 for(int i=originalsize-1;i>=0;i--) {
13 result.add(addNumber + result.get(i));
14 }
15 return result;
16 }