给定一个 BCD(二进制编码的十进制)数,任务是将 BCD 数转换为其等效的十进制数。
例子:
Input: BCD = 100000101000
Output: 828
Explanation:
Dividing the number into chunks of 4, it becomes 1000 0010 1000.
Here, 1000 is equivalent to 8 and
0010 is equivalent to 2.
So, the number becomes 828.
Input: BCD = 1001000
Output: 48
Explanation:
Dividing the number into chunks of 4, it becomes 0100 1000.
Here, 0100 is equivalent to 4 and
1000 is equivalent to 8.
So, the number becomes 48.
方法:
- 迭代给定 BCD 数字中的所有位。
- 将给定的 BCD 数分成 4 个块,并开始计算其等效的Decimal 数。
- 将此数字存储在名为sum的变量中。
- 从存储在变量num中的总和中的数字开始构建一个数字。
- 反转到目前为止形成的数字并返回该数字。
下面是上述方法的实现。
C++
// C++ code to convert BCD to its
// decimal number(base 10).
// Including Header Files
#include
using namespace std;
// Function to convert BCD to Decimal
int bcdToDecimal(string s)
{
int len = s.length(),
check = 0, check0 = 0;
int num = 0, sum = 0,
mul = 1, rev = 0;
// Iterating through the bits backwards
for (int i = len - 1; i >= 0; i--) {
// Forming the equivalent
// digit(0 to 9)
// from the group of 4.
sum += (s[i] - '0') * mul;
mul *= 2;
check++;
// Reinitialize all variables
// and compute the number.
if (check == 4 || i == 0) {
if (sum == 0 && check0 == 0) {
num = 1;
check0 = 1;
}
else {
// update the answer
num = num * 10 + sum;
}
check = 0;
sum = 0;
mul = 1;
}
}
// Reverse the number formed.
while (num > 0) {
rev = rev * 10 + (num % 10);
num /= 10;
}
if (check0 == 1)
return rev - 1;
return rev;
}
// Driver Code
int main()
{
string s = "100000101000";
// Function Call
cout << bcdToDecimal(s);
return 0;
}
Java
// Java code to convert BCD to its
// decimal number(base 10).
// Including Header Files
import java.io.*;
import java.util.*;
class GFG {
// Function to convert BCD to Decimal
public static int bcdToDecimal(String s)
{
int len = s.length();
int check = 0, check0 = 0;
int num = 0, sum = 0;
int mul = 1, rev = 0;
// Iterating through the bits backwards
for(int i = len - 1; i >= 0; i--)
{
// Forming the equivalent
// digit(0 to 9)
// from the group of 4.
sum += (s.charAt(i) - '0') * mul;
mul *= 2;
check++;
// Reinitialize all variables
// and compute the number.
if (check == 4 || i == 0)
{
if (sum == 0 && check0 == 0)
{
num = 1;
check0 = 1;
}
else
{
// Update the answer
num = num * 10 + sum;
}
check = 0;
sum = 0;
mul = 1;
}
}
// Reverse the number formed.
while (num > 0)
{
rev = rev * 10 + (num % 10);
num /= 10;
}
if (check0 == 1)
return rev - 1;
return rev;
}
// Driver code
public static void main(String[] args)
{
String s = "100000101000";
// Function Call
System.out.println(bcdToDecimal(s));
}
}
// This code is contributed by coder001
Python3
# Python3 code to convert BCD to its
# decimal number(base 10).
# Function to convert BCD to Decimal
def bcdToDecimal(s):
length = len(s);
check = 0;
check0 = 0;
num = 0;
sum = 0;
mul = 1;
rev = 0;
# Iterating through the bits backwards
for i in range(length - 1, -1, -1):
# Forming the equivalent
# digit(0 to 9)
# from the group of 4.
sum += (ord(s[i]) - ord('0')) * mul;
mul *= 2;
check += 1;
# Reinitialize all variables
# and compute the number
if (check == 4 or i == 0):
if (sum == 0 and check0 == 0):
num = 1;
check0 = 1;
else:
# Update the answer
num = num * 10 + sum;
check = 0;
sum = 0;
mul = 1;
# Reverse the number formed.
while (num > 0):
rev = rev * 10 + (num % 10);
num //= 10;
if (check0 == 1):
return rev - 1;
return rev;
# Driver Code
if __name__ == "__main__":
s = "100000101000";
# Function Call
print(bcdToDecimal(s));
# This code is contributed by AnkitRai01
C#
// C# code to convert BCD to its
// decimal number(base 10).
// Including Header Files
using System;
class GFG
{
// Function to convert BCD to Decimal
public static int bcdToDecimal(String s)
{
int len = s.Length;
int check = 0, check0 = 0;
int num = 0, sum = 0;
int mul = 1, rev = 0;
// Iterating through the bits backwards
for(int i = len - 1; i >= 0; i--)
{
// Forming the equivalent
// digit(0 to 9)
// from the group of 4.
sum += (s[i] - '0') * mul;
mul *= 2;
check++;
// Reinitialize all variables
// and compute the number.
if (check == 4 || i == 0)
{
if (sum == 0 && check0 == 0)
{
num = 1;
check0 = 1;
}
else
{
// Update the answer
num = num * 10 + sum;
}
check = 0;
sum = 0;
mul = 1;
}
}
// Reverse the number formed.
while (num > 0)
{
rev = rev * 10 + (num % 10);
num /= 10;
}
if (check0 == 1)
return rev - 1;
return rev;
}
// Driver code
public static void Main(String[] args)
{
String s = "100000101000";
// Function Call
Console.WriteLine(bcdToDecimal(s));
}
}
// This code is contributed by 29AjayKumar
Javascript
输出:
828
时间复杂度: O(N)
辅助空间复杂度: O(1)
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