89. 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.

The idea from case n - 1 to n is:

1.First Half : place 0  to all the n - 1 number, 

2.Second Half: place 1 to all the n - 1 number,

3.Reverse second half.

4.Join together

take 2 -> 3 as example

00 01 10 11

1.000 001 010 011 

2.100 101 110 111

3. 111 110 101 100

4. 000 001 010 011 111 110 101 100

public class Solution {
    
    
    public List<Integer> grayCode(int n) {
        if(n == 0){
            List<Integer> ret = new ArrayList();
            ret.add(0);
            return ret;
        }
        
        List<Integer> prev = grayCode(n - 1);
        List<Integer> ret = new ArrayList(prev);
        Collections.reverse(prev);
        int toAdd = 1 << (n - 1);
        for(int i = 0 ; i < prev.size(); i++){
            prev.set(i,prev.get(i) + toAdd);
        }
        ret.addAll(prev);
        return ret;
    }
}

Another solution:

public List<Integer> grayCode(int n) {
    List<Integer> result = new LinkedList<>();
    for (int i = 0; i < 1<<n; i++) result.add(i ^ i>>1);
    return result;
}

000 ^ 000= 000

001 ^ 000 = 001

010 ^ 001 = 011

011 ^ 001 = 010

100 ^ 010 = 110

101 ^ 010 = 111

110 ^ 011 = 101

111 ^ 011 = 100