2014
11-18

# LeetCode-Gray Code[数组]

### 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.

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

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

\begin{center}
\includegraphics[width=160pt]{gray-code-construction.png}\\
\figcaption{The first few steps of the reflect-and-prefix method.}\label{fig:gray-code-construction}
\end{center}

// LeetCode, 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<int> grayCode(int n) {
vector<int> result;
const size_t size = 1 << n;  // 2^n
result.reserve(size);
for (size_t i = 0; i < size; ++i)
result.push_back(binary_to_gray(i));
return result;
}
private:
static unsigned int binary_to_gray(unsigned int n) {
return n ^ (n >> 1);
}
};

// LeetCode, 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<int> grayCode(int n) {
vector<int> result;
result.reserve(1<<n);
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;
}
};

1. 其实国内大部分公司对算法都不够重视。特别是中小型公司老板根本都不懂技术，也不懂什么是算法，从而也不要求程序员懂什么算法，做程序从来不考虑性能问题，只要页面能显示出来就是好程序，这是国内的现状，很无奈。