# 89. Gray Code (Medium)

https://leetcode.com/problems/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.

Example 1:

```Input: 2
Output: `[0,1,3,2]`
Explanation:
00 - 0
01 - 1
11 - 3
10 - 2

For a given n, a gray code sequence may not be uniquely defined.
For example, [0,2,3,1] is also a valid gray code sequence.

00 - 0
10 - 2
11 - 3
01 - 1
```

Example 2:

```Input: 0
Output: ```[0]
Explanation: We define the gray code sequence to begin with 0.
A gray code sequence of n has size = 2n, which for n = 0 the size is 20 = 1.
Therefore, for n = 0 the gray code sequence is [0].```
```

## Solutions

``````class Solution {
public List<Integer> grayCode(int n) {
List<Integer> ans = new ArrayList<>();
for (int i = 0; i < (1 << n); i++) {
}

// assume x is even, x=?0, y=x+1, then y=?1, odd, so that x^y=0[0,k]1
// assume x is odd, x=?01[1,j], y=x+1, then y=?10[1,j], even, so that x^y=0[0,k]11[1,j]

// assume i-1 is is even, such that i is odd and (i-1)>>1 = i>>1, i, i-1 oly differ in the last bit.
// then ans.get(i)^ans.get(i-1)=i^(i>>1)^(i-1)^((i-1)>>1)=i^(i-1)^(i>>1)^((i-1)>>1)=1^0=1
// we can see the only difference is the last bit

// assume i-1 is is odd, such that i is even and (i-1)>>1 = (i>>1)-1
// then ans.get(i)^ans.get(i-1)=i^(i>>1)^(i-1)^((i-1)>>1)=i^(i-1)^(i>>1)^((i-1)>>1)=11[1,k]^01[1,k]=10[1,k]
// we can see the only difference is the first bit

return ans;
}
}
``````