格雷碼生成算法

在博客中看到過一次格雷碼生成算法，我在這裏也想寫一下。

原文中的算法爲：假設已經生成了k位格雷碼，那麼k+1位格雷碼的生成方式爲(1) 按序在k位格雷碼前插入一位0，生成一組編碼，(2)按逆序在k位格雷碼前插入一位1，生成另一組編碼，兩組編碼合起來就是k+1位格雷碼。

以下例：

已有2位格雷碼：00, 01, 11, 10，要生成3位格雷碼，採用此算法：

(1)按序在各碼前插入0，生成000,001, 011,010；

(2)按逆序在各碼前插入1，生成110,111, 101,100；

(3)將兩組編碼組合起來：000, 001, 011, 010, 110, 111, 101, 100，爲3位格雷碼。

 

另一種算法與此算法相似，不一樣的是插入的位是在格雷碼的後面：

對於k位格雷碼，在各格雷碼後面分別插入0, 1 或 1, 0，生成兩個編碼，全部插入完成後組合起來的編碼爲k+1位格雷碼。

如已有2位格雷碼：00,01,11,10，生成3位格雷碼，採用此算法：

(1)在00編碼後面分別插入0,1，生成000, 001;

(2)在01編碼後面分別插入1,0，生成011, 010;

(3)在11編碼後面分別插入0,1，生成110, 111;

(4)在10編碼後面分別插入1,0，生成101,100;

(5)將生成的編碼組合起來：000, 001, 011, 010, 110, 111, 101, 100，爲3位格雷碼。

 

#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;
}

運行時間的測試：

12位格雷碼，方法一和方法二所需時鐘數

 

16位格雷碼，兩種方法所需時鐘數