[LeetCode] Gray Code (Java)

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

1
2
3
4

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.

Analysis

To solve this problem, we need to look at the gray code sequence. Assuming n = 3, we have:

000
001
011
010
110
111
101
100

We can see that the last two digits of 4 codes at the bottom is just the descending sequence of the first 4 codes. The first 4 codes are 0, 1, 3, 2. So, we can easily get the last 4 codes: 2 + 4, 3 + 4, 1 + 4, 0 + 4, which is 6, 7, 5, 4. We can keep doing this until we reach n digits.

Code

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

public class Solution {

public List<Integer> grayCode(int n) {

List<Integer> ret = new LinkedList<>();

ret.add(0);

for (int i = 0; i < n; i++) {

int size = ret.size();

for (int j = size - 1; j >= 0; j--)

ret.add(ret.get(j) + size);

}

return ret;

}

Complexity

We need to generate $2^n$ numbers, so the complexity is $O(2^n)$.