将二进制数转换为八进制

问题是将给定的二进制数(以字符串表示)转换为其等效的八进制数。输入可能非常大，甚至可能无法放入unsigned long long int中。

例子：

Input : 110001110
Output : 616

Input  : 1111001010010100001.010110110011011
Output : 1712241.26633 

想法是将二进制输入视为字符串字符，然后执行以下步骤：

1. 获取小数点左边和右边的子字符串的长度，分别为left_len和right_len 。
2. 如果left_len不是3的倍数，请在开头添加最小数0，以使left子字符串的长度为3的倍数。
3. 如果right_len不是3的倍数，请在末尾添加最小数0，以使right子字符串的长度是3的倍数。
4. 现在，从左侧提取长度为3的一个一的子字符串，并将其相应的八进制代码添加到结果中。
5. 如果在十进制(’。’)之间遇到，则将其添加到结果中。

C++

// C++ implementation to convert a binary number
// to octal number
#include 
using namespace std;
  
// function to create map between binary
// number and its equivalent octal
void createMap(unordered_map *um)
{
    (*um)["000"] = '0';
    (*um)["001"] = '1';
    (*um)["010"] = '2';
    (*um)["011"] = '3';
    (*um)["100"] = '4';
    (*um)["101"] = '5';
    (*um)["110"] = '6';
    (*um)["111"] = '7';   
}
  
// Function to find octal equivalent of binary
string convertBinToOct(string bin)
{
    int l = bin.size();
    int t = bin.find_first_of('.');
      
    // length of string before '.'
    int len_left = t != -1 ? t : l;
      
    // add min 0's in the beginning to make
    // left substring length divisible by 3
    for (int i = 1; i <= (3 - len_left % 3) % 3; i++)
        bin = '0' + bin;
      
    // if decimal point exists   
    if (t != -1)   
    {
        // length of string after '.'
        int len_right = l - len_left - 1;
          
        // add min 0's in the end to make right
        // substring length divisible by 3
        for (int i = 1; i <= (3 - len_right % 3) % 3; i++)
            bin = bin + '0';
    }
      
    // create map between binary and its
    // equivalent octal code
    unordered_map bin_oct_map;
    createMap(&bin_oct_map);
      
    int i = 0;
    string octal = "";
      
    while (1)
    {
        // one by one extract from left, substring
        // of size 3 and add its octal code
        octal += bin_oct_map[bin.substr(i, 3)];
        i += 3;
        if (i == bin.size())
            break;
              
        // if '.' is encountered add it to result
        if (bin.at(i) == '.')   
        {
            octal += '.';
            i++;
        }
    }
      
    // required octal number
    return octal;   
}
  
// Driver program to test above
int main()
{
    string bin = "1111001010010100001.010110110011011";
    cout << "Octal number = "
         << convertBinToOct(bin);
    return 0;    
}

Java

// Java implementation to convert a
// binary number to octal number
import java.io.*;
import java.util.*;
 
class GFG{
 
// Function to create map between binary
// number and its equivalent hexadecimal
static void createMap(Map um)
{
    um.put("000", '0');
    um.put("001", '1');
    um.put("010", '2');
    um.put("011", '3');
    um.put("100", '4');
    um.put("101", '5');
    um.put("110", '6');
    um.put("111", '7');
}
 
// Function to find octal equivalent of binary
static String convertBinToOct(String bin)
{
    int l = bin.length();
    int t = bin.indexOf('.');
 
    // Length of string before '.'
    int len_left = t != -1 ? t : l;
 
    // Add min 0's in the beginning to make
    // left substring length divisible by 3
    for(int i = 1;
            i <= (3 - len_left % 3) % 3;
            i++)
        bin = '0' + bin;
 
    // If decimal point exists
    if (t != -1)
    {
         
        // Length of string after '.'
        int len_right = l - len_left - 1;
 
        // add min 0's in the end to make right
        // substring length divisible by 3
        for(int i = 1;
                i <= (3 - len_right % 3) % 3;
                i++)
            bin = bin + '0';
    }
 
    // Create map between binary and its
    // equivalent octal code
    Map bin_oct_map = new HashMap();
    createMap(bin_oct_map);
 
    int i = 0;
    String octal = "";
 
    while (true)
    {
         
        // One by one extract from left, substring
        // of size 3 and add its octal code
        octal += bin_oct_map.get(
            bin.substring(i, i + 3));
        i += 3;
         
        if (i == bin.length())
            break;
 
        // If '.' is encountered add it to result
        if (bin.charAt(i) == '.')
        {
            octal += '.';
            i++;
        }
    }
 
    // Required octal number
    return octal;
}
 
// Driver code
public static void main(String[] args)
{
    String bin = "1111001010010100001.010110110011011";
    System.out.println("Octal number = " +
                        convertBinToOct(bin));
}
}
 
// This code is contributed by jithin

Python3

# Python3 implementation to convert a binary number
# to octal number
 
# function to create map between binary
# number and its equivalent octal
def createMap(bin_oct_map):
    bin_oct_map["000"] = '0'
    bin_oct_map["001"] = '1'
    bin_oct_map["010"] = '2'
    bin_oct_map["011"] = '3'
    bin_oct_map["100"] = '4'
    bin_oct_map["101"] = '5'
    bin_oct_map["110"] = '6'
    bin_oct_map["111"] = '7'
 
# Function to find octal equivalent of binary
def convertBinToOct(bin):
    l = len(bin)
     
    # length of string before '.'
    t = -1
    if '.' in bin:
        t = bin.index('.')
        len_left = t
    else:
        len_left = l
     
    # add min 0's in the beginning to make
    # left substring length divisible by 3
    for i in range(1, (3 - len_left % 3) % 3 + 1):
        bin = '0' + bin
     
    # if decimal point exists
    if (t != -1):
         
        # length of string after '.'
        len_right = l - len_left - 1
         
        # add min 0's in the end to make right
        # substring length divisible by 3
        for i in range(1, (3 - len_right % 3) % 3 + 1):
            bin = bin + '0'
     
    # create dictionary between binary and its
    # equivalent octal code
    bin_oct_map = {}
    createMap(bin_oct_map)
    i = 0
    octal = ""
     
    while (True) :
         
        # one by one extract from left, substring
        # of size 3 and add its octal code
        octal += bin_oct_map[bin[i:i + 3]]
        i += 3
        if (i == len(bin)):
            break
             
        # if '.' is encountered add it to result
        if (bin[i] == '.'):
            octal += '.'
            i += 1
             
    # required octal number
    return octal
 
# Driver Code
bin = "1111001010010100001.010110110011011"
print("Octal number = ",
       convertBinToOct(bin))
 
# This code is contributed
# by Atul_kumar_Shrivastava

输出：

Octal number = 1712241.26633

时间复杂度：O(n)，其中n是字符串的长度。