LeetCode89 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.
本題重點(diǎn)是理解格雷碼的生成規(guī)律锤岸，先試著寫出3位格雷碼：
000 - 0
001 - 1
011 - 3
010 - 2
110 - 6
111 - 7
101 - 5
100 - 4
或
000 - 0
001 - 1
011 - 3
010 - 2
110 - 6
100 - 4
101 - 5
111 - 7
一開始我很糾結(jié)到底是哪種靴患。。。網(wǎng)上查閱+自己對(duì)于2位的情況后课蔬，確定應(yīng)該使用第一種！：岜ぁ玻靡！
然后看0132與6754
發(fā)現(xiàn)誒？６章Ｉ颐А！
6754-4444=2310
0132與2310正好是逆序＜榕；杳！
有沒有很神奇阵面？Ｇ峋帧洪鸭！
ok
了解這點(diǎn)以后，可以采用回溯仑扑，對(duì)于n位greycode：
1 先dfs(n-1)獲取n-1位greycode
2 將n-1位的greycode都加入n位的list中
3 將n-1位的值都加上2^(n-1)次方的offset后览爵，再加入list中
public class Solution {
public List<Integer> grayCode(int n) {
List<Integer> codes = new ArrayList<>();
dfs(n, codes);
return codes;
}
public void dfs(int n, List<Integer> codes) {
if (n == 0) {
codes.add(0);
return;
}
dfs(n-1, codes);
int offset = (int)Math.pow(2,n-1);
int size = codes.size()-1;
for (int i = size; i >= 0; i--) {
codes.add(offset + codes.get(i));
}
}
}
