计数要翻转的位数，将 A 转换为 B | Set-2
原文:https://www . geesforgeks . org/count-要翻转的位数-转换-a 到 b-set-2/
给定两个整数 A 和 B ，任务是计算将 A 转换为 B 所需翻转的位数。 举例:
输入: A = 10，B = 7 输出: 3 二进制(10) = 1010 二进制(7)= 0111 1010 0111 3 位需要翻转。 输入: A = 8，B = 7 输出: 4
方法:解决这个问题的方法已经讨论过了这里。这里，需要翻转的位数可以通过逐个匹配两个整数中的所有位来找到。如果所考虑的位不同，则递增计数。 以下是上述方法的实施:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the count of bits
// to be flipped to convert a to b
int countBits(int a, int b)
{
// To store the required count
int count = 0;
// Loop until both of them become zero
while (a || b) {
// Store the last bits in a
// as well as b
int last_bit_a = a & 1;
int last_bit_b = b & 1;
// If the current bit is not same
// in both the integers
if (last_bit_a != last_bit_b)
count++;
// Right shift both the integers by 1
a = a >> 1;
b = b >> 1;
}
// Return the count
return count;
}
// Driver code
int main()
{
int a = 10, b = 7;
cout << countBits(a, b);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the count of bits
// to be flipped to convert a to b
static int countBits(int a, int b)
{
// To store the required count
int count = 0;
// Loop until both of them become zero
while (a > 0 || b > 0)
{
// Store the last bits in a
// as well as b
int last_bit_a = a & 1;
int last_bit_b = b & 1;
// If the current bit is not same
// in both the integers
if (last_bit_a != last_bit_b)
count++;
// Right shift both the integers by 1
a = a >> 1;
b = b >> 1;
}
// Return the count
return count;
}
// Driver code
public static void main(String[] args)
{
int a = 10, b = 7;
System.out.println(countBits(a, b));
}
}
// This code is contributed by Princi Singh
Python 3
# Python3 implementation of the approach
# Function to return the count of bits
# to be flipped to convert a to b
def countBits(a, b):
# To store the required count
count = 0
# Loop until both of them become zero
while (a or b):
# Store the last bits in a
# as well as b
last_bit_a = a & 1
last_bit_b = b & 1
# If the current bit is not same
# in both the integers
if (last_bit_a != last_bit_b):
count += 1
# Right shift both the integers by 1
a = a >> 1
b = b >> 1
# Return the count
return count
# Driver code
a = 10
b = 7
print(countBits(a, b))
# This code is contributed by Mohit Kumar
C
// C# implementation of the above approach
using System;
class GFG
{
// Function to return the count of bits
// to be flipped to convert a to b
static int countBits(int a, int b)
{
// To store the required count
int count = 0;
// Loop until both of them become zero
while (a > 0 || b > 0)
{
// Store the last bits in a
// as well as b
int last_bit_a = a & 1;
int last_bit_b = b & 1;
// If the current bit is not same
// in both the integers
if (last_bit_a != last_bit_b)
count++;
// Right shift both the integers by 1
a = a >> 1;
b = b >> 1;
}
// Return the count
return count;
}
// Driver code
public static void Main(String[] args)
{
int a = 10, b = 7;
Console.WriteLine(countBits(a, b));
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the count of bits
// to be flipped to convert a to b
function countBits(a, b)
{
// To store the required count
var count = 0;
// Loop until both of them become zero
while (a || b) {
// Store the last bits in a
// as well as b
var last_bit_a = a & 1;
var last_bit_b = b & 1;
// If the current bit is not same
// in both the integers
if (last_bit_a != last_bit_b)
count++;
// Right shift both the integers by 1
a = a >> 1;
b = b >> 1;
}
// Return the count
return count;
}
// Driver code
var a = 10, b = 7;
document.write(countBits(a, b));
</script>
Output:
3
时间复杂度:O(最小值(对数 a，对数 b))
辅助空间:0(1)
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