通过交换不相等的位对来最大化给定整数

给定正整数N ，任务是确定可以通过对给定整数N执行以下操作来形成的最大可能整数：

• 将整数转换为其二进制表示形式。
• 仅以二进制表示形式交换不相等的位。

例子：

Input : 11
Output : 14
Explanation : 

• (11)₁₀ = (1011)₂
• Swap 0^th bit with 2^nd bit to get(1110)₂ = (14)₁₀

Input : 9
Output : 12

方法：按照给定的步骤解决问题

• 计算数字N中的设置位数。
• 在任何二进制字符串，使用上述给定的操作，所有设置的位都可以移到左侧，所有未设置的位都可以移到右侧。这是因为任何一对不等比特(0，1)可以交换，即，“0110110”可以使用3个操作转换为“1111000”
• 因为，通过将所有置位的比特向左和所有未置位的比特向右分离，可以使给定的整数最大化。

下面是上述方法的实现：

C++

// C++ implementation
// of the above approach
#include 
using namespace std;
 
// Function to return the maximum possible
// value that can be obtained from the
// given integer by performing given operations
int findMaxNum(int num)
{
   
    // Convert to equivalent
    // binary representation
    bitset<4> b(num);
     
    string binaryNumber = b.to_string();
 
    // Stores binary representation
    // of the maximized value
    string maxBinaryNumber = "";
 
    // Store the count of 0's and 1's
    int count0 = 0, count1 = 0;
 
    // Stores the total Number of Bits
    int N = 4;
 
    for (int i = 0; i < N; i++) {
 
        // If current bit is set
        if (binaryNumber[i] == '1') {
 
            // Increment Count1
            count1++;
        }
        else {
 
            // Increment Count0
            count0++;
        }
    }
 
    // Shift all set bits to left
    // to maximize the value
    for (int i = 0; i < count1; i++) {
        maxBinaryNumber += '1';
    }
 
    // Shift all unset bits to right
    // to maximize the value
    for (int i = 0; i < count0; i++) {
        maxBinaryNumber += '0';
    }
     
    // Return the maximized value
    return stoi(maxBinaryNumber, 0, 2);
}
 
// Driver code
int main()
{
    // Given "Input
    int N = 11;
  
    // Function call to find the
    // Maximum Possible Number
    cout << findMaxNum(N);
    return 0;
}
 
// This code is contributed by divyeshrabadiya07.

Java

// Java implementation
// of the above approach
 
import java.io.*;
 
class GFG {
 
    // Function to return the maximum possible
    // value that can be obtained from the
    // given integer by performing given operations
    public static int findMaxNum(int num)
    {
        // Convert to equivalent
        // binary representation
        String binaryNumber
            = Integer.toBinaryString(num);
 
        // Stores binary representation
        // of the maximized value
        String maxBinaryNumber = "";
 
        // Store the count of 0's and 1's
        int count0 = 0, count1 = 0;
 
        // Stores the total Number of Bits
        int N = binaryNumber.length();
 
        for (int i = 0; i < N; i++) {
 
            // If current bit is set
            if (binaryNumber.charAt(i) == '1') {
 
                // Increment Count1
                count1++;
            }
            else {
 
                // Increment Count0
                count0++;
            }
        }
 
        // Shift all set bits to left
        // to maximize the value
        for (int i = 0; i < count1; i++) {
            maxBinaryNumber += '1';
        }
 
        // Shift all unset bits to right
        // to maximize the value
        for (int i = 0; i < count0; i++) {
            maxBinaryNumber += '0';
        }
 
        // Return the maximized value
        return Integer.parseInt(
            maxBinaryNumber, 2);
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        // Given "Input
        int N = 11;
 
        // Function call to find the
        // Maximum Possible Number
        System.out.println(findMaxNum(N));
    }
}

Python3

# Python3 implementation
# of the above approach
 
# Function to return the maximum possible
# value that can be obtained from the
# given integer by performing given operations
def findMaxNum(num):
   
    # Convert to equivalent
    # binary representation
    binaryNumber = bin(num)[2:]
 
    # Stores binary representation
    # of the maximized value
    maxBinaryNumber = ""
 
    # Store the count of 0's and 1's
    count0, count1 = 0, 0
 
    # Stores the total Number of Bits
    N = len(binaryNumber)
    for i in range(N):
 
        # If current bit is set
        if (binaryNumber[i] == '1'):
 
            # Increment Count1
            count1 += 1
        else:
 
            # Increment Count0
            count0 += 1
 
    # Shift all set bits to left
    # to maximize the value
    for i in range(count1):
        maxBinaryNumber += '1'
 
    # Shift all unset bits to right
    # to maximize the value
    for i in range(count0):
        maxBinaryNumber += '0'
 
    # Return the maximized value
    return int(maxBinaryNumber, 2)
 
# Driver Code
if __name__ == '__main__':
     
    # Given "Input
    N = 11
 
    # Function call to find the
    # Maximum Possible Number
    print(findMaxNum(N))
     
    # This code is contributed by mohit kumar 29.

C#

// C# implementation
// of the above approach
using System;
public class GFG
{
 
  // Function to return the maximum possible
  // value that can be obtained from the
  // given integer by performing given operations
  public static int findMaxNum(int num)
  {
 
    // Convert to equivalent
    // binary representation
    string binaryNumber =  Convert.ToString(num, 2);
 
    // Stores binary representation
    // of the maximized value
    string maxBinaryNumber = "";
 
    // Store the count of 0's and 1's
    int count0 = 0, count1 = 0;
 
    // Stores the total Number of Bits
    int N = binaryNumber.Length;
    for (int i = 0; i < N; i++)
    {
 
      // If current bit is set
      if (binaryNumber[i] == '1')
      {
 
        // Increment Count1
        count1++;
      }
      else
      {
 
        // Increment Count0
        count0++;
      }
    }
 
    // Shift all set bits to left
    // to maximize the value
    for (int i = 0; i < count1; i++)
    {
      maxBinaryNumber += '1';
    }
 
    // Shift all unset bits to right
    // to maximize the value
    for (int i = 0; i < count0; i++)
    {
      maxBinaryNumber += '0';
    }
 
    // Return the maximized value
    return Convert.ToInt32(maxBinaryNumber, 2);
  }
 
  // Driver Code
  static public void Main()
  {
 
    // Given "Input
    int N = 11;
 
    // Function call to find the
    // Maximum Possible Number
    Console.WriteLine(findMaxNum(N));
  }
}
 
// This code is contributed by Dharanendra L V

Javascript

 
// javascript implementation
// of the above approach
 
// Function to return the maximum possible
// value that can be obtained from the
// given integer by performing given operations
function findMaxNum(num)
{
    // Convert to equivalent
    // binary representation
    var binaryNumber
        = Number(num).toString(2);;
    // Stores binary representation
    // of the maximized value
    var maxBinaryNumber = "";
 
    // Store the count of 0's and 1's
    var count0 = 0, count1 = 0;
 
    // Stores the total Number of Bits
    var N = binaryNumber.length;
 
    for (i = 0; i < N; i++) {
 
        // If current bit is set
        if (binaryNumber.charAt(i) == '1') {
 
            // Increment Count1
            count1++;
        }
        else {
 
            // Increment Count0
            count0++;
        }
    }
 
    // Shift all set bits to left
    // to maximize the value
    for (i = 0; i < count1; i++) {
        maxBinaryNumber += '1';
    }
 
    // Shift all unset bits to right
    // to maximize the value
    for (i = 0; i < count0; i++) {
        maxBinaryNumber += '0';
    }
 
    // Return the maximized value
    return parseInt(maxBinaryNumber,2);
}
 
// Driver Code
 
 // Given "Input
var N = 11;
 
// Function call to find the
// Maximum Possible Number
document.write(findMaxNum(N));
 
// This code contributed by Princi Singh
 

输出：

14

时间复杂度： O(log(N))
辅助空间： O(1)