leetcode 89. Gray Code

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本文介绍了一种使用递归方法生成灰码序列的算法。灰码是一种特殊的二进制数序列，相邻两个数值仅有一位不同。文章详细解释了递归生成过程，并通过实例展示了如何从一位灰码扩展到多位。

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:

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.

递归生成码表

这种方法基于格雷码是反射码的事实，利用递归的如下规则来构造：

1位格雷码有两个码字
(n+1)位格雷码中的前2n个码字等于n位格雷码的码字，按顺序书写，加前缀0
(n+1)位格雷码中的后2n个码字等于n位格雷码的码字，按逆序书写，加前缀1[3]
n+1位格雷码的集合 = n位格雷码集合(顺序)加前缀0 + n位格雷码集合(逆序)加前缀1

2位格雷码	3位格雷码	4位格雷码	4位自然二进制码
00 01 11 10	000 001 011 010 110 111 101 100	0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000	0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111

class Solution {
	vector<vector<int>> getGray(int n) {
		vector<vector<int>>re;
		if(n==0)
			{
			vector<int>aa;
			aa.push_back(0);
			re.push_back(aa);
			return re;
		}
		if (n == 1)
		{
			vector<int>aa, bb;
			aa.push_back(0);
			bb.push_back(1);
			re.push_back(aa); re.push_back(bb);
			return re;
		}
		vector<vector<int>>rr = getGray(n - 1);
		for (int i = 0; i < rr.size(); i++)
		{
			vector<int>ss = rr[i];
			ss.insert(ss.begin(), 0);
			re.push_back(ss);
		}
		for (int i = rr.size() - 1; i >= 0; i--)
		{
			vector<int>ss = rr[i];
			ss.insert(ss.begin(), 1);
			re.push_back(ss);
			1 << 2;
		}
		return re;
	}
public:
	vector<int> grayCode(int n) {
		vector<vector<int>>rr = getGray(n);
		vector<int>re;
		for (int i = 0; i < rr.size(); i++)
		{
			int dd = 0;
			for (int j = 0; j < n; j++)
				dd += (rr[i][j] << (n - 1 - j));
			re.push_back(dd);
		}
		return re;
	}
};