## Gray Code

### 描述

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.

### 分析

**自然二进制码转换为格雷码**：\$\$g_0=b_0, g_i=b_i \oplus b_{i-1}\$\$

**格雷码转换为自然二进制码**：\$\$b_0=g_0, b_i=g_i \oplus b_{i-1}\$\$

![The first few steps of the reflect-and-prefix method.](../images/gray-code-construction.png)

### 数学公式

{% if book.java %}
```java
// Gray Code
// 数学公式，时间复杂度O(2^n)，空间复杂度O(1)
public class Solution {
public ArrayList grayCode(int n) {
final int size = 1 << n; // 2^n
ArrayList result = new ArrayList<>(size);

for (int i = 0; i < size; ++i)
return result;
}
private static int binary_to_gray(int n) {
return n ^ (n >> 1);
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Gray Code
// 数学公式，时间复杂度O(2^n)，空间复杂度O(1)
class Solution {
public:
vector grayCode(int n) {
const int size = 1 << n; // 2^n
vector result;
result.reserve(size);

for (int i = 0; i < size; ++i)
result.push_back(binary_to_gray(i));
return result;
}
private:
static int binary_to_gray(int n) {
return n ^ (n >> 1);
}
};
```
{% endif %}

### Reflect-and-prefix method

{% if book.java %}
```java
// Gray Code
// reflect-and-prefix method
// 时间复杂度O(2^n)，空间复杂度O(1)
public class Solution {
public ArrayList grayCode(int n) {
final int size = 1 << n;
ArrayList result = new ArrayList<>(size);

for (int i = 0; i < n; i++) {
final int highest_bit = 1 << i;
for (int j = result.size() - 1; j >= 0; j--) // 要反着遍历，才能对称
}
return result;
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Gray Code
// reflect-and-prefix method
// 时间复杂度O(2^n)，空间复杂度O(1)
class Solution {
public:
vector grayCode(int n) {
const int size = 1 << n;
vector result;
result.reserve(size);

result.push_back(0);
for (int i = 0; i < n; i++) {
const int highest_bit = 1 << i;
for (int j = result.size() - 1; j >= 0; j--) // 要反着遍历，才能对称
result.push_back(highest_bit | result[j]);
}
return result;
}
};
```
{% endif %}