Leetcode: 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.

Understand of the problem:
The problem asks for outputting all Gray Code. For the background knowledge about the gray code, please refer the link 
http://zh.wikipedia.org/wiki/%E6%A0%BC%E9%9B%B7%E7%A0%81#.E4.BA.8C.E9.80.B2.E4.BD.8D.E6.95.B8.E8.BD.89.E6.A0.BC.E9.9B.B7.E7.A2.BC

For n = 1's Gray Code:
0
---
1

For n = 2 Gray Code:
00
01
-----
11
10

For n = 3 Gray Code:
000
001
011
010
----------
110
111
101
100

So we can see that for the Gray code sequence with n, it equals to (n - 1)'s Gray code sequence, as shown in the upper half the examples. The below halves are equal to the sum of n- 1 gray code plus the 1 << (n - 1), in backward order. 

Code (Java):

public class Solution {
    public List<Integer> grayCode(int n) {
        List<Integer> result = new ArrayList<Integer>();
        
        if (n == 0) {
            result.add(0);
            return result;
        }
        
        List<Integer> lastGray = grayCode(n - 1);
        int addOnNum = 1 << (n - 1);
        
        result.addAll(lastGray);
        
        for (int i = lastGray.size() - 1; i >= 0; i--) {
            result.add(lastGray.get(i) + addOnNum);
        }
        
        return result;
    }
}

Update on 8/1/2018:

class Solution {
    public List<Integer> grayCode(int n) {
        List<Integer> ans = new ArrayList<>();
        
        if (n <= 0) {
            ans.add(0);
            return ans;
        }
        
        ans.add(0);
        ans.add(1);
        
        for (int i = 2; i <= n; i++) {
            int size = ans.size();
            for (int j = size - 1; j >= 0; j--) {
                ans.add(ans.get(j) + (1 <<(i - 1)));
            }
        }
        
        return ans;
    }
}