# 算法设计：生成n位灰度码 |set 2

2021年3月10日16:01:18 发表评论 344 次浏览

## 本文概述

``````Input: n=2
Output: 00 01 11 10
Every adjacent element of gray code differs
only by one bit.
So the n bit grey codes are: 00 01 11 10

Input: n=3
Output: 000 001 011 010 110 111 101 100
Every adjacent element of gray code differs
only by one bit.
So the n bit gray codes are:
000 001 011 010 110 111 101 100``````

## 推荐：请尝试以下方法{IDE}首先, 在继续解决方案之前。

1. 灰度码的第一位(MSB)与二进制数的第一位(MSB)相同。
2. 灰度码的第二位(从左侧开始)等于二进制数的第一位(MSB)与第二位(2nd MSB)的XOR。
3. 灰度码的第三位(从左侧开始)等于第二位(第二MSB)和第三位(第三MSB)的XOR, 依此类推。

## C ++

``````// C++ program to generate n-bit
// gray codes
#include <bits/stdc++.h>
using namespace std;

// Function to convert decimal to binary
void decimalToBinaryNumber( int x, int n)
{
int * binaryNumber = new int (x);
int i = 0;
while (x > 0) {
binaryNumber[i] = x % 2;
x = x / 2;
i++;
}

// leftmost digits are filled with 0
for ( int j = 0; j < n - i; j++)
cout << '0' ;

for ( int j = i - 1; j >= 0; j--)
cout << binaryNumber[j];
}

// Function to generate gray code
void generateGrayarr( int n)
{
int N = 1 << n;
for ( int i = 0; i < N; i++) {

// generate gray code of corresponding
// binary number of integer i.
int x = i ^ (i >> 1);

// printing gray code
decimalToBinaryNumber(x, n);

cout << endl;
}
}

// Drivers code
int main()
{
int n = 3;
generateGrayarr(n);
return 0;
}``````

## Java

``````// Java program to generate
// n-bit gray codes
import java.io.*;

class GFG {

// Function to convert
// decimal to binary
static void decimalToBinaryNumber( int x, int n)
{
int [] binaryNumber = new int [x];
int i = 0 ;
while (x > 0 ) {
binaryNumber[i] = x % 2 ;
x = x / 2 ;
i++;
}

// leftmost digits are
// filled with 0
for ( int j = 0 ; j < n - i; j++)
System.out.print( '0' );

for ( int j = i - 1 ;
j >= 0 ; j--)
System.out.print(binaryNumber[j]);
}

// Function to generate
// gray code
static void generateGrayarr( int n)
{
int N = 1 << n;
for ( int i = 0 ; i < N; i++) {

// generate gray code of
// corresponding binary
// number of integer i.
int x = i ^ (i >> 1 );

// printing gray code
decimalToBinaryNumber(x, n);

System.out.println();
}
}

// Driver code
public static void main(String[] args)
{
int n = 3 ;
generateGrayarr(n);
}
}

// This code is contributed
// by anuj_67.``````

## Python3

``````# Python program to generate
# n-bit gray codes

# Function to convert
# decimal to binary
def decimalToBinaryNumber(x, n):
binaryNumber = [ 0 ] * x;
i = 0 ;
while (x > 0 ):
binaryNumber[i] = x % 2 ;
x = x / / 2 ;
i + = 1 ;

# leftmost digits are
# filled with 0
for j in range ( 0 , n - i):
print ( '0' , end = "");

for j in range (i - 1 , - 1 , - 1 ):
print (binaryNumber[j], end = "");

# Function to generate
# gray code
def generateGrayarr(n):
N = 1 << n;
for i in range (N):

# generate gray code of
# corresponding binary
# number of integer i.
x = i ^ (i >> 1 );

# printing gray code
decimalToBinaryNumber(x, n);

print ();

# Driver code
if __name__ = = '__main__' :
n = 3 ;
generateGrayarr(n);

# This code is contributed by 29AjayKumar``````

## C#

``````// C# program to generate
// n-bit gray codes
using System;

class GFG {

// Function to convert
// decimal to binary
static void decimalToBinaryNumber( int x, int n)
{
int [] binaryNumber = new int [x];
int i = 0;
while (x > 0) {
binaryNumber[i] = x % 2;
x = x / 2;
i++;
}

// leftmost digits are
// filled with 0
for ( int j = 0; j < n - i; j++)
Console.Write( '0' );

for ( int j = i - 1;
j >= 0; j--)
Console.Write(binaryNumber[j]);
}

// Function to generate
// gray code
static void generateGrayarr( int n)
{
int N = 1 << n;
for ( int i = 0; i < N; i++) {

// Generate gray code of
// corresponding binary
// number of integer i.
int x = i ^ (i >> 1);

// printing gray code
decimalToBinaryNumber(x, n);

Console.WriteLine();
}
}

// Driver code
public static void Main()
{
int n = 3;
generateGrayarr(n);
}
}

// This code is contributed
// by anuj_67.``````

``````000
001
011
010
110
111
101
100``````

• 时间复杂度：上)。
从0到(n-1)只需一个遍历。
• 辅助空间：O(对数x)。
(x)的二进制表示需要空格(log x)。