两种n位格雷码生成算法

在博客中看到过一次格雷码生成算法，我在这里也想写一下。
原文中的算法为：假设已经生成了k位格雷码，那么k+1位格雷码的生成方式为(1) 按序在k位格雷码前插入一位0，生成一组编码，(2)按逆序在k位格雷码前插入一位1，生成另外一组编码，两组编码合起来就是k+1位格雷码。
如下例：
已有2位格雷码：00, 01, 11, 10，要生成3位格雷码，采用此算法：
(1)按序在各码前插入0，生成000,001,
011,010；
(2)按逆序在各码前插入1，生成110,111,
101,100；
(3)将两组编码组合起来：000, 001, 011, 010, 110, 111, 101, 100，为3位格雷码。

另外一种算法与此算法类似，不同的是插入的位是在格雷码的后面：
对于k位格雷码，在各格雷码后面分别插入0, 1 或 1, 0，生成两个编码，所有插入完成后组合起来的编码为k+1位格雷码。
如已有2位格雷码：00,01,11,10，生成3位格雷码，采用此算法：
(1)在00编码后面分别插入0,1，生成000, 001;
(2)在01编码后面分别插入1,0，生成011, 010;
(3)在11编码后面分别插入0,1，生成110, 111;
(4)在10编码后面分别插入1,0，生成101,100;
(5)将生成的编码组合起来：000, 001, 011, 010, 110, 111, 101, 100，为3位格雷码。

代码如下：

#include <iostream>
#include <vector>
#include <string>
#include <time.h>

void GrayCodeOne(int num);
void GrayCodeTwo(int num);

using namespace std;

int main()
{
int count;
cout << "Input Code Number:";
cin >> count;

cout << "Produce Gray Code using method 1" << endl;
clock_t beginOne = clock();
GrayCodeOne(count);
clock_t endOne = clock();
cout << "Gray Code First Method using time: " << (endOne - beginOne) << endl;

cout << "Produce Gray Code using method 2" << endl;
clock_t beginTwo = clock();
GrayCodeTwo(count);
clock_t endTwo = clock();
cout << "Gray Code Second Method using time: " << (endTwo - beginTwo) << endl;

return 0;
}

// Method to produce gray code using method inserting 0 in front of old gray code by positive
// and inserting 1 in front of old gray code by nagative.
void GrayCodeOne(int num)
{
if (num < 1)
{
cout << "Error input Integer" << endl;
return;
}

vector<string> codeVec;

int cIdx = 1;
for (; cIdx <= num; cIdx++)
{
if (codeVec.size() < 2)
{
codeVec.push_back("0");
codeVec.push_back("1");
}
else
{
vector<string> tranVec;
tranVec.resize(2 * codeVec.size());
int tranIdx = 0;
vector<string>::iterator codeIter = codeVec.begin();
for (; codeIter != codeVec.end(); codeIter++)
{
string str = "0";
str.append(*codeIter);
tranVec[tranIdx++] = str;
}

vector<string>::reverse_iterator rCodeIter = codeVec.rbegin();
for (; rCodeIter != codeVec.rend(); rCodeIter++)
{
string str = "1";
str.append(*rCodeIter);
tranVec[tranIdx++] = str;
}

codeVec.assign(tranVec.begin(), tranVec.end());
}
}

//vector<string>::iterator vecIter = codeVec.begin();
//for (; vecIter != codeVec.end(); vecIter++)
//{
//	cout << *vecIter << endl;
//}

return;
}

// Method to produce gray code using method inserting 0/1 in the back of first gray code
// then inserting 1/0 in the back of next gray code.
void GrayCodeTwo(int num)
{
if (num < 1)
{
cout << "Input error Integer" << endl;
return;
}

vector<string> codeVec;

int cIdx = 1;
for (; cIdx <= num; cIdx++)
{
if (codeVec.size() < 2)
{
codeVec.push_back("0");
codeVec.push_back("1");
}
else
{
vector<string> tranVec;
int tranIdx = 0;
int cIdx = codeVec.size();

tranVec.resize(2 * cIdx);
for (int vIdx = 0; vIdx < cIdx; vIdx++)
{
string str = codeVec[vIdx];
if (0 == (vIdx % 2))
{
string str0 = str;
str0.append("0");
tranVec[tranIdx++] = str0;

string str1 = str;
str1.append("1");
tranVec[tranIdx++] = str1;
}
else
{
string str0 = str;
str0.append("1");
tranVec[tranIdx++] = str0;

string str1 = str;
str1.append("0");
tranVec[tranIdx++] = str1;
}
}

codeVec.assign(tranVec.begin(), tranVec.end());
}
}

//vector<string>::iterator vecIter = codeVec.begin();
//for (; vecIter != codeVec.end(); vecIter++)
//{
//	cout << *vecIter << endl;
//}

return;
}

运行时间的测试：
12位格雷码，方法一和方法二所需时钟数



16位格雷码，两种方法所需时钟数