# 下一个更大数字的二进制表示，具有相同的1和0

2021年3月24日14:36:33 发表评论 705 次浏览

## 本文概述

``````Input : 10010
Output : 10100
Here n = (18)10 = (10010)2
next greater = (20)10 = (10100)2
Binary representation of 20 contains same number of
1's and 0's as in 18.

Input : 111000011100111110
Output :  111000011101001111``````

## 推荐：请尝试以下方法{IDE}首先, 在继续解决方案之前。

1. 遍历二进制字符串bstr从右边。
2. 遍历时找到第一个索引一世这样bstr [i] ='0'和bstr [i + 1] ='1'。
3. 在索引" i"和" i + 1"处交换字符。
4. 由于我们需要最小的下一个值, 因此请考虑索引中的子字符串我+2结束并移动所有1个在最后的子字符串中。

## C ++

``````// C++ program to find next permutation in a
// binary string.
#include <bits/stdc++.h>
using namespace std;

// Function to find the next greater number
// with same number of 1's and 0's
string nextGreaterWithSameDigits(string bnum)
{
int l = bnum.size();
int i;
for ( int i=l-2; i>=1; i--)
{
// locate first 'i' from end such that
// bnum[i]=='0' and bnum[i+1]=='1'
// swap these value and break;
if (bnum.at(i) == '0' &&
bnum.at(i+1) == '1' )
{
char ch = bnum.at(i);
bnum.at(i) = bnum.at(i+1);
bnum.at(i+1) = ch;
break ;
}
}

// if no swapping performed
if (i == 0)
"no greater number" ;

// Since we want the smallest next value, // shift all 1's at the end in the binary
// substring starting from index 'i+2'
int j = i+2, k = l-1;
while (j < k)
{
if (bnum.at(j) == '1' && bnum.at(k) == '0' )
{
char ch = bnum.at(j);
bnum.at(j) = bnum.at(k);
bnum.at(k) = ch;
j++;
k--;
}

// special case while swapping if '0'
// occurs then break
else if (bnum.at(i) == '0' )
break ;

else
j++;

}

// required next greater number
return bnum;
}

// Driver program to test above
int main()
{
string bnum = "10010" ;
cout << "Binary representation of next greater number = "
<< nextGreaterWithSameDigits(bnum);
return 0;
}``````

## Java

``````// Java program to find next permutation in a
// binary string.
class GFG
{

// Function to find the next greater number
// with same number of 1's and 0's
static String nextGreaterWithSameDigits( char [] bnum)
{
int l = bnum.length;
int i;
for (i = l - 2 ; i >= 1 ; i--)
{
// locate first 'i' from end such that
// bnum[i]=='0' and bnum[i+1]=='1'
// swap these value and break;
if (bnum[i] == '0' &&
bnum[i+ 1 ] == '1' )
{
char ch = bnum[i];
bnum[i] = bnum[i+ 1 ];
bnum[i+ 1 ] = ch;
break ;
}
}

// if no swapping performed
if (i == 0 )
System.out.println( "no greater number" );

// Since we want the smallest next value, // shift all 1's at the end in the binary
// substring starting from index 'i+2'
int j = i + 2 , k = l - 1 ;
while (j < k)
{
if (bnum[j] == '1' && bnum[k] == '0' )
{
char ch = bnum[j];
bnum[j] = bnum[k];
bnum[k] = ch;
j++;
k--;
}

// special case while swapping if '0'
// occurs then break
else if (bnum[i] == '0' )
break ;

else
j++;

}

// required next greater number
return String.valueOf(bnum);
}

// Driver program to test above
public static void main(String[] args)
{
char [] bnum = "10010" .toCharArray();
System.out.println( "Binary representation of next greater number = "
+ nextGreaterWithSameDigits(bnum));
}
}

// This code contributed by Rajput-Ji``````

## Python3

``````# Python3 program to find next permutation in a
# binary string.

# Function to find the next greater number
# with same number of 1's and 0's
def nextGreaterWithSameDigits(bnum):
l = len (bnum)
bnum = list (bnum)
for i in range (l - 2 , 0 , - 1 ):

# locate first 'i' from end such that
# bnum[i]=='0' and bnum[i+1]=='1'
# swap these value and break
if (bnum[i] = = '0' and bnum[i + 1 ] = = '1' ):
ch = bnum[i]
bnum[i] = bnum[i + 1 ]
bnum[i + 1 ] = ch
break

# if no swapping performed
if (i = = 0 ):
return "no greater number"

# Since we want the smallest next value, # shift all 1's at the end in the binary
# substring starting from index 'i+2'
j = i + 2
k = l - 1
while (j < k):
if (bnum[j] = = '1' and bnum[k] = = '0' ):
ch = bnum[j]
bnum[j] = bnum[k]
bnum[k] = ch
j + = 1
k - = 1

# special case while swapping if '0'
# occurs then break
elif (bnum[i] = = '0' ):
break
else :
j + = 1

# required next greater number
return bnum

# Driver code
bnum = "10010"
print ( "Binary representation of next greater number = " , * nextGreaterWithSameDigits(bnum), sep = "")

# This code is contributed by shubhamsingh10``````

## C#

``````// C# program to find next permutation in a
// binary string.
using System;

class GFG
{

// Function to find the next greater number
// with same number of 1's and 0's
static String nextGreaterWithSameDigits( char [] bnum)
{
int l = bnum.Length;
int i;
for (i = l - 2; i >= 1; i--)
{
// locate first 'i' from end such that
// bnum[i]=='0' and bnum[i+1]=='1'
// swap these value and break;
if (bnum[i] == '0' &&
bnum[i+1] == '1' )
{
char ch = bnum[i];
bnum[i] = bnum[i+1];
bnum[i+1] = ch;
break ;
}
}

// if no swapping performed
if (i == 0)
Console.WriteLine( "no greater number" );

// Since we want the smallest next value, // shift all 1's at the end in the binary
// substring starting from index 'i+2'
int j = i + 2, k = l - 1;
while (j < k)
{
if (bnum[j] == '1' && bnum[k] == '0' )
{
char ch = bnum[j];
bnum[j] = bnum[k];
bnum[k] = ch;
j++;
k--;
}

// special case while swapping if '0'
// occurs then break
else if (bnum[i] == '0' )
break ;

else
j++;

}

// required next greater number
return String.Join( "" , bnum);
}

// Driver code
public static void Main(String[] args)
{
char [] bnum = "10010" .ToCharArray();
Console.WriteLine( "Binary representation of next greater number = "
+ nextGreaterWithSameDigits(bnum));
}
}

// This code is contributed by 29AjayKumar``````

``Binary representation of next greater number = 10100``