本文概述
给定数字n, 请生成从0到2 ^ n-1的位模式, 以使连续的模式相差一位。
例子:
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。
另一种方法生成n位灰度码已经讨论过了。
方法:
想法是使用XOR和Right shift操作获得二进制数的灰度码。
- 灰度码的第一位(MSB)与二进制数的第一位(MSB)相同。
- 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。
- 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。
这样, 可以为相应的二进制数计算灰度码。因此, 可以观察到, 第i个元素可以由i和floor(i / 2)的按位XOR形成, 等于i和(i >> 1)的按位XOR, 即i右移1。通过执行此操作, 二进制数的MSB保持不变, 所有其他位与相邻的更高位按位进行XOR。
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.
输出如下:
000
001
011
010
110
111
101
100
复杂度分析:
- 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)。