求 N 和 N 的 32 位二进制表示反转形成的数之间的最大值
原文:https://www . geeksforgeeks . org/find-n 和 n 的 32 位二进制表示反转形成的数字之间的最大值/
给定一个正的 32 位整数 N ，任务是找出 N 的值与十进制表示对求反得到的数之间的最大值NT11】在一个 32 位整数中的二进制表示。
示例:
输入: N = 6 输出: 1610612736 解释: 6 在 32 位整数中的二进制表示为 00000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 反转这个二进制字符串得到(01100000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000 N 与得到的数字之间的最大值为 1610612736。所以，打印 1610612736。
输入:N = 1610612736 T3】输出: 1610612736
方法:按照以下步骤解决问题:
- 计算数字 N 的二进制字符串，并将其存储在字符串 S 中。
- 反转弦 S 。
- 计算变量中新反转字符串 S 的十进制值，比如 M 。
- 完成上述步骤后，打印 N 和 M 的最大值作为结果。
下面是上述方法的实现:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function that obtains the number
// using said operations from N
int reverseBin(int N)
{
// Stores the binary representation
// of the number N
string S = "";
int i;
// Find the binary representation
// of the number N
for (i = 0; i < 32; i++) {
// Check for the set bits
if (N & (1LL << i))
S += '1';
else
S += '0';
}
// Reverse the string S
reverse(S.begin(), S.end());
// Stores the obtained number
int M = 0;
// Calculating the decimal value
for (i = 0; i < 32; i++) {
// Check for set bits
if (S[i] == '1')
M += (1LL << i);
}
return M;
}
// Function to find the maximum value
// between N and the obtained number
int maximumOfTwo(int N)
{
int M = reverseBin(N);
return max(N, M);
}
// Driver Code
int main()
{
int N = 6;
cout << maximumOfTwo(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function that obtains the number
// using said operations from N
static int reverseBin(int N)
{
// Stores the binary representation
// of the number N
String S = "";
int i;
// Find the binary representation
// of the number N
for (i = 0; i < 32; i++) {
// Check for the set bits
if ((N & (1L << i)) != 0)
S += '1';
else
S += '0';
}
// Reverse the string S
S = (new StringBuilder(S)).reverse().toString();
// Stores the obtained number
int M = 0;
// Calculating the decimal value
for (i = 0; i < 32; i++) {
// Check for set bits
if (S.charAt(i) == '1')
M += (1L << i);
}
return M;
}
// Function to find the maximum value
// between N and the obtained number
static int maximumOfTwo(int N)
{
int M = reverseBin(N);
return Math.max(N, M);
}
// Driver Code
public static void main(String[] args)
{
int N = 6;
System.out.print(maximumOfTwo(N));
}
}
// This code is contributed by Kingash.
Python 3
# Python3 program for the above approach
# Function that obtains the number
# using said operations from N
def reverseBin(N):
# Stores the binary representation
# of the number N
S = ""
i = 0
# Find the binary representation
# of the number N
for i in range(32):
# Check for the set bits
if (N & (1 << i)):
S += '1'
else:
S += '0'
# Reverse the string S
S = list(S)
S = S[::-1]
S = ''.join(S)
# Stores the obtained number
M = 0
# Calculating the decimal value
for i in range(32):
# Check for set bits
if (S[i] == '1'):
M += (1 << i)
return M
# Function to find the maximum value
# between N and the obtained number
def maximumOfTwo(N):
M = reverseBin(N)
return max(N, M)
# Driver Code
if __name__ == '__main__':
N = 6
print(maximumOfTwo(N))
# This code is contributed by SURENDRA_GANGWAR
C
// C# program for the above approach
using System;
class GFG{
static string ReverseString(string s)
{
char[] array = s.ToCharArray();
Array.Reverse(array);
return new string(array);
}
// Function that obtains the number
// using said operations from N
public static int reverseBin(int N)
{
// Stores the binary representation
// of the number N
string S = "";
int i;
// Find the binary representation
// of the number N
for (i = 0; i < 32; i++) {
// Check for the set bits
if ((N & (1L << i)) != 0)
S += '1';
else
S += '0';
}
// Reverse the string S
S = ReverseString(S);
// Stores the obtained number
int M = 0;
// Calculating the decimal value
for (i = 0; i < 32; i++) {
// Check for set bits
if (S[i] == '1')
M += (1 << i);
}
return M;
}
// Function to find the maximum value
// between N and the obtained number
static int maximumOfTwo(int N)
{
int M = reverseBin(N);
return Math.Max(N, M);
}
// Driver Code
static void Main()
{
int N = 6;
Console.Write(maximumOfTwo(N));
}
}
// This code is contributed by SoumikMondal
java 描述语言
<script>
// JavaScript program for the above approach
// Function that obtains the number
// using said operations from N
function reverseBin(N)
{
// Stores the binary representation
// of the number N
let S = "";
let i;
// Find the binary representation
// of the number N
for (i = 0; i < 32; i++) {
// Check for the set bits
if (N & (1 << i))
S += '1';
else
S += '0';
}
// Reverse the string S
S = S.split("").reverse().join("");
// Stores the obtained number
let M = 0;
// Calculating the decimal value
for (i = 0; i < 32; i++) {
// Check for set bits
if (S[i] == '1')
M += (1 << i);
}
return M;
}
// Function to find the maximum value
// between N and the obtained number
function maximumOfTwo(N)
{
let M = reverseBin(N);
return Math.max(N, M);
}
// Driver Code
let N = 6;
document.write(maximumOfTwo(N));
</script>
Output:
1610612736
时间复杂度:O(32) T5辅助空间:** O(1)
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